A Dart Is Thrown Horizontally

4 Motility in Two and Three Dimensions

iv.3 Projectile Motion

Learning Objectives

By the stop of this section, you volition exist able to:

• Use one-dimensional move in perpendicular directions to analyze projectile motion.
• Calculate the range, time of flight, and maximum height of a projectile that is launched and impacts a flat, horizontal surface.
• Discover the time of flying and bear on velocity of a projectile that lands at a different peak from that of launch.
• Calculate the trajectory of a projectile.

Projectile move is the movement of an object thrown or projected into the air, subject area only to acceleration as a issue of gravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth's atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their path is chosen a trajectory. The motion of falling objects as discussed in Motion Along a Directly Line is a unproblematic one-dimensional type of projectile movement in which there is no horizontal movement. In this department, nosotros consider two-dimensional projectile motion, and our treatment neglects the effects of air resistance.

The well-nigh of import fact to think hither is that motions along perpendicular axes are independent and thus can be analyzed separately. We discussed this fact in Displacement and Velocity Vectors, where we saw that vertical and horizontal motions are independent. The key to analyzing 2-dimensional projectile motion is to break information technology into 2 motions: ane along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible because acceleration resulting from gravity is vertical; thus, in that location is no acceleration forth the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the ten-axis and the vertical axis the y-axis. Information technology is non required that nosotros use this choice of axes; it is simply user-friendly in the example of gravitational acceleration. In other cases we may cull a unlike set of axes. Effigy illustrates the notation for displacement, where we define [latex]\mathbf{\overset{\to }{s}}[/latex] to be the total displacement, and [latex]\mathbf{\overset{\to }{x}}[/latex] and [latex]\mathbf{\overset{\to }{y}}[/latex] are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are southward, 10, and y.

Figure iv.11 The total displacement due south of a soccer ball at a bespeak along its path. The vector [latex]\mathbf{\overset{\to }{s}}[/latex] has components [latex]\mathbf{\overset{\to }{x}}[/latex] and [latex]\mathbf{\overset{\to }{y}}[/latex] along the horizontal and vertical axes. Its magnitude is s and it makes an angle θ with the horizontal.

To describe projectile move completely, we must include velocity and dispatch, every bit well as displacement. We must find their components along the x- and y-axes. Let's assume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be upwardly, the components of acceleration are and then very simple:

[latex]{a}_{y}=\text{−}yard=-nine.8\,\text{m}\text{/}{\text{southward}}^{ii}\enspace(-32\,\text{ft}\text{/}{\text{s}}^{2}).[/latex]

Considering gravity is vertical, [latex]{a}_{x}=0.[/latex] If [latex]{a}_{x}=0,[/latex] this means the initial velocity in the x direction is equal to the final velocity in the ten management, or [latex]{5}_{x}={v}_{0x}.[/latex] With these conditions on dispatch and velocity, nosotros can write the kinematic (Equation) through (Equation) for motion in a uniform gravitational field, including the rest of the kinematic equations for a constant dispatch from Motion with Constant Dispatch. The kinematic equations for movement in a uniform gravitational field become kinematic equations with [latex]{a}_{y}=\text{−}yard,\enspace{a}_{x}=0:[/latex]

Horizontal Movement

[latex]{v}_{0x}={5}_{x},\,x={x}_{0}+{five}_{x}t[/latex]

Vertical Motion

[latex]y={y}_{0}+\frac{i}{two}({v}_{0y}+{v}_{y})t[/latex]

[latex]{v}_{y}={5}_{0y}-gt[/latex]

[latex]y={y}_{0}+{5}_{0y}t-\frac{1}{2}g{t}^{2}[/latex]

[latex]{v}_{y}^{2}={five}_{0y}^{2}-2g(y-{y}_{0})[/latex]

Using this set of equations, we can analyze projectile movement, keeping in mind some important points.

Problem-Solving Strategy: Projectile Motion

1. Resolve the motion into horizontal and vertical components along the x– and y-axes. The magnitudes of the components of deportation [latex]\mathbf{\overset{\to }{s}}[/latex] forth these axes are 10 and y. The magnitudes of the components of velocity [latex]\mathbf{\overset{\to }{v}}[/latex] are [latex]{v}_{x}=five\text{cos}\,\theta \,\text{and}\,{v}_{y}=5\text{sin}\,\theta ,[/latex] where 5 is the magnitude of the velocity and θ is its direction relative to the horizontal, as shown in Figure.
2. Treat the motion as two contained ane-dimensional motions: ane horizontal and the other vertical. Use the kinematic equations for horizontal and vertical motion presented earlier.
3. Solve for the unknowns in the two split motions: one horizontal and one vertical. Note that the only common variable between the motions is fourth dimension t. The problem-solving procedures here are the same as those for one-dimensional kinematics and are illustrated in the following solved examples.
4. Recombine quantities in the horizontal and vertical directions to find the total displacement [latex]\mathbf{\overset{\to }{s}}[/latex] and velocity [latex]\mathbf{\overset{\to }{v}}.[/latex] Solve for the magnitude and direction of the displacement and velocity using

   [latex]southward=\sqrt{{x}^{2}+{y}^{2}},\enspace\theta ={\text{tan}}^{-1}(y\text{/}ten),\enspace{v}=\sqrt{{five}_{x}^{2}+{v}_{y}^{2}},[/latex]

   where θ is the direction of the displacement [latex]\mathbf{\overset{\to }{s}}.[/latex]

Effigy four.12 (a) We analyze two-dimensional projectile movement by breaking it into two contained one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is unproblematic, because [latex]{a}_{x}=0[/latex] and [latex]{v}_{10}[/latex] is a constant. (c) The velocity in the vertical direction begins to decrease every bit the object rises. At its highest point, the vertical velocity is nothing. As the object falls toward Earth again, the vertical velocity increases again in magnitude but points in the opposite management to the initial vertical velocity. (d) The x and y motions are recombined to give the total velocity at any given point on the trajectory.

Example

A Fireworks Projectile Explodes High and Away

During a fireworks display, a crush is shot into the air with an initial speed of 70.0 thou/due south at an bending of [latex]75.0^\circ[/latex] higher up the horizontal, as illustrated in Effigy. The fuse is timed to ignite the shell just every bit information technology reaches its highest point above the basis. (a) Calculate the height at which the trounce explodes. (b) How much time passes between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? (d) What is the total displacement from the point of launch to the highest point?

Figure 4.13 The trajectory of a fireworks vanquish. The fuse is prepare to explode the vanquish at the highest point in its trajectory, which is constitute to be at a height of 233 m and 125 thousand away horizontally.

Strategy

The motion can be broken into horizontal and vertical motions in which [latex]{a}_{x}=0[/latex] and [latex]{a}_{y}=\text{−}yard.[/latex] We can then define [latex]{x}_{0}[/latex] and [latex]{y}_{0}[/latex] to exist zilch and solve for the desired quantities.

Solution

(a) By "peak" nosotros mean the altitude or vertical position y above the starting betoken. The highest point in any trajectory, called the apex, is reached when [latex]{five}_{y}=0.[/latex] Since nosotros know the initial and terminal velocities, besides as the initial position, we utilize the following equation to observe y:

[latex]{5}_{y}^{2}={v}_{0y}^{ii}-2g(y-{y}_{0}).[/latex]

Considering [latex]{y}_{0}[/latex] and [latex]{v}_{y}[/latex] are both zero, the equation simplifies to

[latex]\text{0}={5}_{0y}^{2}-2gy.[/latex]

Solving for y gives

[latex]y=\frac{{five}_{0y}^{2}}{2g}.[/latex]

Now we must observe [latex]{5}_{0y},[/latex] the component of the initial velocity in the y management. Information technology is given past [latex]{v}_{0y}={v}_{0}\text{sin}{\theta }_{0},[/latex] where [latex]{five}_{0}[/latex] is the initial velocity of 70.0 1000/s and [latex]{\theta }_{0}=75^\circ[/latex] is the initial angle. Thus,

[latex]{v}_{0y}={v}_{0}\text{sin}\,\theta =(lxx.0\,\text{one thousand}\text{/}\text{s})\text{sin}\,75^\circ=67.vi\,\text{m}\text{/}\text{south}[/latex]

and y is

[latex]y=\frac{{(67.half-dozen\,\text{grand}\text{/}\text{southward})}^{2}}{2(nine.80\,\text{m}\text{/}{\text{s}}^{2})}.[/latex]

Thus, we take

[latex]y=233\,\text{one thousand}\text{.}[/latex]

Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, just the acceleration resulting from gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that whatever projectile with a 67.6-m/southward initial vertical component of velocity reaches a maximum height of 233 k (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In do, air resistance is not completely negligible, and then the initial velocity would have to be somewhat larger than that given to achieve the same summit.

(b) As in many physics bug, there is more than ane manner to solve for the time the projectile reaches its highest indicate. In this case, the easiest method is to utilize [latex]{v}_{y}={five}_{0y}-gt.[/latex] Because [latex]{five}_{y}=0[/latex] at the apex, this equation reduces to simply

[latex]0={v}_{0y}-gt[/latex]

or

[latex]t=\frac{{v}_{0y}}{thousand}=\frac{67.six\,\text{m}\text{/}\text{southward}}{nine.80\,\text{k}\text{/}{\text{s}}^{2}}=6.90\text{s}\text{.}[/latex]

This fourth dimension is also reasonable for big fireworks. If yous are able to see the launch of fireworks, notice that several seconds pass before the trounce explodes. Another way of finding the time is by using[latex]y\,\text{=}\,{y}_{0}+\frac{i}{2}({five}_{0y}+{v}_{y})t.[/latex] This is left for you as an do to complete.

(c) Because air resistance is negligible, [latex]{a}_{ten}=0[/latex] and the horizontal velocity is abiding, as discussed earlier. The horizontal deportation is the horizontal velocity multiplied by time as given by [latex]ten={x}_{0}+{v}_{x}t,[/latex] where [latex]{10}_{0}[/latex] is equal to zero. Thus,

[latex]x={5}_{ten}t,[/latex]

where [latex]{5}_{x}[/latex] is the x-component of the velocity, which is given past

[latex]{5}_{x}={v}_{0}\text{cos}\,\theta =(lxx.0\,\text{grand}\text{/}\text{s})\text{cos}75^\circ=eighteen.i\,\text{m}\text{/}\text{s}.[/latex]

Time t for both motions is the aforementioned, and so ten is

[latex]x=(18.one\,\text{grand}\text{/}\text{s})6.90\,\text{southward}=125\,\text{yard}\text{.}[/latex]

Horizontal movement is a abiding velocity in the absence of air resistance. The horizontal displacement found hither could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, air resistance has a major issue, and many fragments land directly below.

(d) The horizontal and vertical components of the displacement were just calculated, so all that is needed hither is to discover the magnitude and management of the deportation at the highest point:

[latex]\mathbf{\overset{\to }{s}}=125\mathbf{\hat{i}}+233\mathbf{\hat{j}}[/latex]

[latex]|\mathbf{\overset{\to }{s}}|=\sqrt{{125}^{2}+{233}^{two}}=264\,\text{m}[/latex]

[latex]\theta ={\text{tan}}^{-1}(\frac{233}{125})=61.8^\circ.[/latex]

Note that the angle for the displacement vector is less than the initial angle of launch. To see why this is, review Figure, which shows the curvature of the trajectory toward the ground level.

When solving Effigy(a), the expression we found for y is valid for whatever projectile motion when air resistance is negligible. Phone call the maximum height y = h. Then,

[latex]h=\frac{{v}_{0y}^{2}}{2g}.[/latex]

This equation defines the maximum height of a projectile above its launch position and information technology depends only on the vertical component of the initial velocity.

Bank check Your Understanding

A rock is thrown horizontally off a cliff [latex]100.0\,\text{chiliad}[/latex] high with a velocity of 15.0 m/southward. (a) Define the origin of the coordinate organization. (b) Which equation describes the horizontal motion? (c) Which equations describe the vertical movement? (d) What is the rock's velocity at the point of impact?

Show Solution

(a) Choose the top of the cliff where the stone is thrown from the origin of the coordinate system. Although it is arbitrary, we typically cull fourth dimension t = 0 to correspond to the origin. (b) The equation that describes the horizontal move is [latex]x={ten}_{0}+{v}_{10}t.[/latex] With [latex]{x}_{0}=0,[/latex] this equation becomes [latex]x={five}_{x}t.[/latex] (c) Figure through Figure and Figure depict the vertical movement, but since [latex]{y}_{0}=0\,\text{and}\,{v}_{0y}=0,[/latex] these equations simplify profoundly to become [latex]y=\frac{ane}{2}({v}_{0y}+{v}_{y})t=\frac{i}{2}{v}_{y}t,\enspace[/latex] [latex]{v}_{y}=\text{−}gt,\enspace[/latex] [latex]y=-\frac{i}{ii}m{t}^{2},\enspace[/latex] and [latex]{five}_{y}^{two}=-2gy.[/latex] (d) We use the kinematic equations to detect the ten and y components of the velocity at the bespeak of impact. Using [latex]{v}_{y}^{two}=-2gy[/latex] and noting the betoken of impact is −100.0 m, we find the y component of the velocity at impact is [latex]{v}_{y}=44.3\,\text{m}\text{/}\text{s}.[/latex] Nosotros are given the ten component, [latex]{v}_{x}=xv.0\,\text{g}\text{/}\text{southward},[/latex] so we can calculate the total velocity at touch on: v = 46.eight m/s and [latex]\theta =71.three^\circ[/latex] below the horizontal.

Example

Calculating Projectile Move: Tennis Actor

A tennis player wins a lucifer at Arthur Ashe stadium and hits a ball into the stands at thirty m/south and at an bending [latex]45^\circ[/latex] higher up the horizontal (Figure). On its fashion downward, the ball is caught by a spectator x m above the point where the ball was hit. (a) Calculate the time it takes the lawn tennis ball to accomplish the spectator. (b) What are the magnitude and management of the brawl'southward velocity at impact?

Figure 4.14 The trajectory of a tennis brawl striking into the stands.

Strategy

Once again, resolving this two-dimensional motion into two independent one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed past its vertical move lonely. Thus, we solve for t outset. While the ball is rising and falling vertically, the horizontal movement continues at a abiding velocity. This example asks for the final velocity. Thus, nosotros recombine the vertical and horizontal results to obtain [latex]\mathbf{\overset{\to }{five}}[/latex] at terminal time t, determined in the first part of the example.

Solution

(a) While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We tin observe the time for this by using Figure:

[latex]y\,\text{=}\,{y}_{0}\,\text{+}\,{v}_{0y}t-\frac{1}{two}g{t}^{ii}.[/latex]

If nosotros take the initial position [latex]{y}_{0}[/latex] to exist zip, then the final position is y = 10 grand. The initial vertical velocity is the vertical component of the initial velocity:

[latex]{v}_{0y}={v}_{0}\text{sin}\,{\theta }_{0}=(thirty.0\,\text{yard}\text{/}\text{due south})\text{sin}\,45^\circ=21.2\,\text{1000}\text{/}\text{s}.[/latex]

Substituting into Figure for y gives us

[latex]10.0\,\text{grand}=(21.2\,\text{m/due south})t-(iv.90\,{\text{one thousand/s}}^{\text{2}}){t}^{ii}.[/latex]

Rearranging terms gives a quadratic equation in t:

[latex](4.ninety\,{\text{thou/due south}}^{\text{2}}){t}^{two}-(21.2\,\text{chiliad/s})t+ten.0\,\text{k}=0.[/latex]

Use of the quadratic formula yields t = 3.79 s and t = 0.54 due south. Since the ball is at a height of 10 m at 2 times during its trajectory—once on the way upwards and one time on the fashion downwardly—we accept the longer solution for the time it takes the brawl to reach the spectator:

[latex]t=three.79\,\text{s}\text{.}[/latex]

The time for projectile motion is determined completely past the vertical motion. Thus, whatever projectile that has an initial vertical velocity of 21.2 g/s and lands 10.0 m below its starting altitude spends iii.79 s in the air.

(b) We tin observe the final horizontal and vertical velocities [latex]{five}_{x}[/latex] and [latex]{5}_{y}[/latex] with the use of the result from (a). So, we can combine them to find the magnitude of the total velocity vector [latex]\mathbf{\overset{\to }{v}}[/latex] and the angle [latex]\theta[/latex] it makes with the horizontal. Since [latex]{v}_{x}[/latex] is constant, nosotros can solve for it at whatever horizontal location. We choose the starting point because we know both the initial velocity and the initial angle. Therefore,

[latex]{v}_{x}={5}_{0}\text{cos}{\theta }_{0}=(30\,\text{m}\text{/}\text{s})\text{cos}\,45^\circ=21.ii\,\text{thou}\text{/}\text{due south}.[/latex]

The final vertical velocity is given by Figure:

[latex]{v}_{y}={v}_{0y}-gt.[/latex]

Since [latex]{five}_{0y}[/latex] was institute in role (a) to be 21.2 grand/due south, we have

[latex]{v}_{y}=21.ii\,\text{yard}\text{/}\text{due south}-9.8\,\text{m}\text{/}{\text{s}}^{2}(3.79\,\text{s})=-15.nine\,\text{m}\text{/}\text{due south}.[/latex]

The magnitude of the final velocity [latex]\mathbf{\overset{\to }{v}}[/latex] is

[latex]v=\sqrt{{v}_{x}^{2}+{5}_{y}^{two}}=\sqrt{{(21.2\,\text{chiliad}\text{/}\text{s})}^{ii}+{(\text{−}\,\text{fifteen}.nine\,\text{m}\text{/}\text{s})}^{2}}=26.5\,\text{one thousand}\text{/}\text{south}.[/latex]

The management [latex]{\theta }_{v}[/latex] is found using the changed tangent:

[latex]{\theta }_{v}={\text{tan}}^{-1}(\frac{{v}_{y}}{{v}_{x}})={\text{tan}}^{-1}(\frac{21.2}{-fifteen.9})=-53.1^\circ.[/latex]

Significance

(a) As mentioned earlier, the fourth dimension for projectile motion is determined completely by the vertical motion. Thus, whatsoever projectile that has an initial vertical velocity of 21.2 m/s and lands ten.0 k below its starting altitude spends three.79 south in the air. (b) The negative angle means the velocity is [latex]53.one^\circ[/latex] below the horizontal at the point of affect. This event is consistent with the fact that the brawl is impacting at a point on the other side of the noon of the trajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than the magnitude of the initial velocity we wait since it is impacting x.0 m above the launch elevation.

Fourth dimension of Flight, Trajectory, and Range

Of interest are the time of flying, trajectory, and range for a projectile launched on a apartment horizontal surface and impacting on the aforementioned surface. In this case, kinematic equations give useful expressions for these quantities, which are derived in the following sections.

Time of flying

We can solve for the time of flying of a projectile that is both launched and impacts on a flat horizontal surface past performing some manipulations of the kinematic equations. Nosotros note the position and displacement in y must be zero at launch and at affect on an even surface. Thus, we set the deportation in y equal to null and find

[latex]y-{y}_{0}={v}_{0y}t-\frac{1}{2}grand{t}^{2}=({v}_{0}\text{sin}{\theta }_{0})t-\frac{1}{ii}g{t}^{2}=0.[/latex]

Factoring, we have

[latex]t({5}_{0}\text{sin}{\theta }_{0}-\frac{gt}{2})=0.[/latex]

Solving for t gives us

[latex]{T}_{\text{tof}}=\frac{2({v}_{0}\text{sin}{\theta }_{0})}{g}.[/latex]

This is the fourth dimension of flight for a projectile both launched and impacting on a flat horizontal surface. Figure does not apply when the projectile lands at a different elevation than it was launched, as we saw in Figure of the tennis histrion hitting the ball into the stands. The other solution, t = 0, corresponds to the fourth dimension at launch. The time of flight is linearly proportional to the initial velocity in the y management and inversely proportional to g. Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the aforementioned velocity as on Globe would be airborne half-dozen times as long.

Trajectory

The trajectory of a projectile can be found by eliminating the fourth dimension variable t from the kinematic equations for capricious t and solving for y(x). We take [latex]{10}_{0}={y}_{0}=0[/latex] and then the projectile is launched from the origin. The kinematic equation for x gives

[latex]x={v}_{0x}t\Rightarrow t=\frac{10}{{v}_{0x}}=\frac{ten}{{v}_{0}\text{cos}{\theta }_{0}}.[/latex]

Substituting the expression for t into the equation for the position [latex]y=({v}_{0}\text{sin}{\theta }_{0})t-\frac{1}{2}yard{t}^{ii}[/latex] gives

[latex]y=({five}_{0}\text{sin}\,{\theta }_{0})(\frac{10}{{v}_{0}\text{cos}\,{\theta }_{0}})-\frac{ane}{2}g{(\frac{ten}{{v}_{0}\text{cos}\,{\theta }_{0}})}^{2}.[/latex]

Rearranging terms, we have

[latex]y=(\text{tan}\,{\theta }_{0})x-[\frac{g}{two{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}]{ten}^{ii}.[/latex]

This trajectory equation is of the form [latex]y=ax+b{10}^{2},[/latex] which is an equation of a parabola with coefficients

[latex]a=\text{tan}\,{\theta }_{0},\enspace b=-\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{two}}.[/latex]

Range

From the trajectory equation we can likewise observe the range, or the horizontal distance traveled by the projectile. Factoring Effigy, we have

[latex]y=10[\text{tan}\,{\theta }_{0}-\frac{g}{two{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}ten].[/latex]

The position y is zero for both the launch point and the touch point, since we are once more considering merely a flat horizontal surface. Setting y = 0 in this equation gives solutions x = 0, corresponding to the launch betoken, and

[latex]x=\frac{2{v}_{0}^{two}\text{sin}\,{\theta }_{0}\text{cos}\,{\theta }_{0}}{g},[/latex]

corresponding to the impact point. Using the trigonometric identity [latex]2\text{sin}\,\theta \text{cos}\,\theta =\text{sin}2\theta[/latex] and setting x = R for range, nosotros find

[latex]R=\frac{{v}_{0}^{two}\text{sin}2{\theta }_{0}}{g}.[/latex]

Note particularly that Figure is valid only for launch and touch on on a horizontal surface. We see the range is straight proportional to the square of the initial speed [latex]{v}_{0}[/latex] and [latex]\text{sin}2{\theta }_{0}[/latex], and it is inversely proportional to the acceleration of gravity. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity. Furthermore, we see from the cistron [latex]\text{sin}2{\theta }_{0}[/latex] that the range is maximum at [latex]45^\circ.[/latex] These results are shown in Figure. In (a) we see that the greater the initial velocity, the greater the range. In (b), we meet that the range is maximum at [latex]45^\circ.[/latex] This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the same range is institute for two initial launch angles that sum to [latex]90^\circ.[/latex] The projectile launched with the smaller angle has a lower apex than the higher angle, but they both have the same range.

Figure four.xv Trajectories of projectiles on level ground. (a) The greater the initial speed [latex]{5}_{0},[/latex] the greater the range for a given initial angle. (b) The effect of initial angle [latex]{\theta }_{0}[/latex] on the range of a projectile with a given initial speed. Note that the range is the same for initial angles of [latex]15^\circ[/latex] and [latex]75^\circ,[/latex] although the maximum heights of those paths are different.

Instance

Comparing Golf Shots

A golfer finds himself in ii different situations on unlike holes. On the second hole he is 120 m from the dark-green and wants to hit the brawl 90 m and let it run onto the green. He angles the shot depression to the ground at [latex]30^\circ[/latex] to the horizontal to let the brawl roll subsequently impact. On the fourth pigsty he is 90 m from the green and wants to let the ball drib with a minimum amount of rolling later on impact. Hither, he angles the shot at [latex]70^\circ[/latex] to the horizontal to minimize rolling after affect. Both shots are hit and impacted on a level surface.

(a) What is the initial speed of the ball at the 2d pigsty?

(b) What is the initial speed of the ball at the fourth hole?

(c) Write the trajectory equation for both cases.

(d) Graph the trajectories.

Strategy

We run across that the range equation has the initial speed and angle, so we tin can solve for the initial speed for both (a) and (b). When we accept the initial speed, we can use this value to write the trajectory equation.

Solution

(a) [latex]R=\frac{{v}_{0}^{2}\text{sin}\,2{\theta }_{0}}{thousand}\Rightarrow {v}_{0}=\sqrt{\frac{Rg}{\text{sin}\,2{\theta }_{0}}}=\sqrt{\frac{xc.0\,\text{m}(9.viii\,\text{m}\text{/}{\text{s}}^{2})}{\text{sin}(2(70^\circ))}}=37.0\,\text{m}\text{/}\text{south}[/latex]

(b) [latex]R=\frac{{v}_{0}^{2}\text{sin}\,2{\theta }_{0}}{thousand}\Rightarrow {v}_{0}=\sqrt{\frac{Rg}{\text{sin}\,2{\theta }_{0}}}=\sqrt{\frac{90.0\,\text{m}(nine.viii\,\text{m}\text{/}{\text{s}}^{2})}{\text{sin}(2(30^\circ))}}=31.9\,\text{m}\text{/}\text{s}[/latex]

(c)

[latex]\begin{assortment}{cc} y=x[\text{tan}\,{\theta }_{0}-\frac{k}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}x]\hfill \\ \text{Second pigsty:}\,y=x[\text{tan}\,lxx^\circ-\frac{9.8\,\text{k}\text{/}{\text{s}}^{2}}{2{[(37.0\,\text{k}\text{/}\text{southward)(}\text{cos}\,70^\circ)]}^{2}}x]=2.75x-0.0306{10}^{2}\hfill \\ \text{Fourth pigsty:}\,y=x[\text{tan}\,30^\circ-\frac{ix.viii\,\text{m}\text{/}{\text{southward}}^{ii}}{2{[(31.9\,\text{m}\text{/}\text{s)(}\text{cos}xxx^\circ)]}^{2}}x]=0.58x-0.0064{x}^{ii}\hfill \terminate{array}[/latex]

(d) Using a graphing utility, we tin can compare the 2 trajectories, which are shown in Effigy.

Effigy iv.16 Two trajectories of a golf brawl with a range of 90 1000. The affect points of both are at the aforementioned level as the launch bespeak.

Significance

The initial speed for the shot at [latex]70^\circ[/latex] is greater than the initial speed of the shot at [latex]30^\circ.[/latex] Notation from Figure that two projectiles launched at the same speed but at unlike angles have the aforementioned range if the launch angles add to [latex]90^\circ.[/latex] The launch angles in this example add to requite a number greater than [latex]xc^\circ.[/latex] Thus, the shot at [latex]70^\circ[/latex] has to have a greater launch speed to reach 90 m, otherwise it would land at a shorter distance.

Check Your Understanding

If the 2 golf shots in Figure were launched at the same speed, which shot would accept the greatest range?

Prove Solution

The golf shot at [latex]thirty^\circ.[/latex]

When we speak of the range of a projectile on level footing, we assume R is very small compared with the circumference of World. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile has farther to fall than it would on level ground, every bit shown in Figure, which is based on a drawing in Newton'due south Principia . If the initial speed is great plenty, the projectile goes into orbit. World'south surface drops 5 m every 8000 m. In 1 southward an object falls 5 chiliad without air resistance. Thus, if an object is given a horizontal velocity of [latex]8000\,\text{one thousand}\text{/}\text{s}[/latex] (or [latex]18,000\text{mi}\text{/}\text{hr})[/latex] near World's surface, it will go into orbit effectually the planet because the surface continuously falls away from the object. This is roughly the speed of the Infinite Shuttle in a depression Earth orbit when it was operational, or any satellite in a low Earth orbit. These and other aspects of orbital motion, such equally World'southward rotation, are covered in greater depth in Gravitation.

Effigy 4.17 Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because Earth curves away beneath its path. With a speed of 8000 m/s, orbit is achieved.

Summary

• Projectile motility is the motion of an object subject field simply to the acceleration of gravity, where the acceleration is constant, equally near the surface of Earth.
• To solve projectile move problems, we analyze the motion of the projectile in the horizontal and vertical directions using the ane-dimensional kinematic equations for x and y.
• The time of flight of a projectile launched with initial vertical velocity [latex]{v}_{0y}[/latex] on an even surface is given by

  [latex]{T}_{tof}=\frac{2({v}_{0}\text{sin}\,\theta )}{1000}.[/latex]

  This equation is valid only when the projectile lands at the same elevation from which it was launched.

• The maximum horizontal altitude traveled by a projectile is called the range. Again, the equation for range is valid just when the projectile lands at the same elevation from which information technology was launched.

Conceptual Questions

Answer the post-obit questions for projectile motion on level basis assuming negligible air resistance, with the initial angle being neither [latex]0^\circ[/latex] nor [latex]ninety^\circ:[/latex] (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever exist the same equally the initial velocity at a time other than at t = 0? (d) Tin the speed ever exist the same as the initial speed at a fourth dimension other than at t = 0?

Testify Solution

a. no; b. minimum at apex of trajectory and maximum at launch and impact; c. no, velocity is a vector; d. yeah, where it lands

Answer the post-obit questions for projectile movement on level footing assuming negligible air resistance, with the initial angle being neither [latex]0^\circ[/latex] nor [latex]xc^\circ:[/latex] (a) Is the dispatch always zero? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the dispatch always reverse in direction to a component of velocity?

A dime is placed at the edge of a tabular array so it hangs over slightly. A quarter is slid horizontally on the table surface perpendicular to the edge and hits the dime caput on. Which coin hits the ground first?

Show Solution

They both striking the ground at the same time.

Problems

A bullet is shot horizontally from shoulder height (one.v m) with and initial speed 200 1000/s. (a) How much time elapses before the bullet hits the footing? (b) How far does the bullet travel horizontally?

Show Solution

a. [latex]t=0.55\,\text{south}[/latex], b. [latex]10=110\,\text{m}[/latex]

A marble rolls off a tabletop 1.0 grand high and hits the floor at a betoken 3.0 grand away from the table'south edge in the horizontal management. (a) How long is the marble in the air? (b) What is the speed of the marble when information technology leaves the table'south border? (c) What is its speed when it hits the floor?

A sprint is thrown horizontally at a speed of 10 k/s at the bull's-eye of a dartboard two.4 m away, equally in the following figure. (a) How far below the intended target does the dart hit? (b) What does your reply tell you virtually how proficient dart players throw their darts?

Evidence Solution

a. [latex]t=0.24\text{s,}\enspace d=0.28\,\text{m}[/latex], b. They aim high.

An airplane flight horizontally with a speed of 500 km/h at a meridian of 800 m drops a crate of supplies (run into the post-obit figure). If the parachute fails to open up, how far in front of the release point does the crate hit the ground?

Suppose the airplane in the preceding problem fires a projectile horizontally in its direction of motion at a speed of 300 grand/due south relative to the plane. (a) How far in front end of the release indicate does the projectile hitting the footing? (b) What is its speed when it hits the basis?

Show Solution

a., [latex]t=12.8\,\text{s,}\enspace x=5619\,\text{m}[/latex] b. [latex]{v}_{y}=125.0\,\text{m}\text{/}\text{s,}\enspace{five}_{10}=439.0\,\text{m}\text{/}\text{s,}\enspace |\mathbf{\overset{\to }{5}}|=456.0\,\text{m}\text{/}\text{due south}[/latex]

A fastball pitcher tin can throw a baseball game at a speed of 40 m/south (xc mi/h). (a) Assuming the pitcher tin release the brawl 16.7 m from home plate so the ball is moving horizontally, how long does it take the ball to reach home plate? (b) How far does the ball drop between the pitcher's mitt and dwelling house plate?

A projectile is launched at an bending of [latex]30^\circ[/latex] and lands 20 due south later at the same height as information technology was launched. (a) What is the initial speed of the projectile? (b) What is the maximum altitude? (c) What is the range? (d) Summate the displacement from the signal of launch to the position on its trajectory at xv s.

Show Solution

a. [latex]{five}_{y}={v}_{0y}-gt,\enspace t=ten\text{south,}\enspace{v}_{y}=0,\enspace{v}_{0y}=98.0\,\text{thousand}\text{/}\text{due south},\enspace{5}_{0}=196.0\,\text{one thousand}\text{/}\text{south}[/latex], b. [latex]h=490.0\,\text{m},[/latex]

c. [latex]{5}_{0x}=169.seven\,\text{chiliad}\text{/}\text{due south,}\enspace 10=3394.0\,\text{m,}[/latex]

d. [latex]\brainstorm{assortment}{cc} x=2545.5\,\text{g}\hfill \\ y=465.5\,\text{grand}\hfill \\ \mathbf{\overset{\to }{south}}=2545.5\,\text{m}\mathbf{\hat{i}}+465.v\,\text{one thousand}\mathbf{\chapeau{j}}\hfill \end{array}[/latex]

A basketball game role player shoots toward a handbasket 6.1 m away and 3.0 thousand above the floor. If the ball is released one.8 m in a higher place the floor at an angle of [latex]60^\circ[/latex] above the horizontal, what must the initial speed exist if it were to go through the basket?

At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of two.0 thou/due south. At this exact instant, a girl throws a brawl horizontally, relative to herself, with an initial speed of 20 m/southward. When she lands, where will she observe the ball? Ignore air resistance.

Show Solution

[latex]-100\,\text{m}=(-two.0\,\text{m}\text{/}\text{s})t-(4.9\,\text{m}\text{/}{\text{south}}^{2}){t}^{2},[/latex] [latex]t=iv.three\,\text{s,}[/latex] [latex]x=86.0\,\text{thousand}[/latex]

A man on a motorcycle traveling at a uniform speed of 10 m/s throws an empty tin can straight upward relative to himself with an initial speed of iii.0 one thousand/southward. Notice the equation of the trajectory equally seen past a law officer on the side of the route. Presume the initial position of the can is the point where it is thrown. Ignore air resistance.

An athlete can jump a altitude of eight.0 m in the broad jump. What is the maximum distance the athlete tin jump on the Moon, where the gravitational acceleration is 1-6th that of Earth?

Show Solution

[latex]{R}_{Moon}=48\,\text{yard}[/latex]

The maximum horizontal altitude a boy can throw a brawl is l m. Assume he can throw with the same initial speed at all angles. How loftier does he throw the ball when he throws it straight upward?

A rock is thrown off a cliff at an angle of [latex]53^\circ[/latex] with respect to the horizontal. The cliff is 100 m loftier. The initial speed of the rock is 30 m/s. (a) How loftier above the edge of the cliff does the rock rise? (b) How far has it moved horizontally when it is at maximum distance? (c) How long after the release does it hit the ground? (d) What is the range of the rock? (e) What are the horizontal and vertical positions of the rock relative to the edge of the cliff at t = 2.0 due south, t = four.0 s, and t = half dozen.0 southward?

Bear witness Solution

a.[latex]{five}_{0y}=24\,\text{yard}\text{/}\text{s}[/latex] [latex]{v}_{y}^{2}={five}_{0y}^{two}-2gy\Rightarrow h=23.4\,\text{thousand}[/latex],

b. [latex]t=iii\,\text{s}\enspace{v}_{0x}=18\,\text{k/southward}\enspace 10=54\,\text{m}[/latex],

c. [latex]y=-100\,\text{m}\enspace{y}_{0}=0[/latex] [latex]y-{y}_{0}={v}_{0y}t-\frac{ane}{two}g{t}^{two}\enspace -100=24t-4.9{t}^{2}[/latex] [latex]\Rightarrow t=seven.58\,\text{due south}[/latex],

d. [latex]x=136.44\,\text{m}[/latex],

e. [latex]t=2.0\,\text{s}\enspace y=28.four\,\text{m}\enspace ten=36\,\text{m}[/latex]

[latex]t=four.0\,\text{south}\enspace y=17.6\,\text{chiliad}\enspace x=22.four\,\text{m}[/latex]

[latex]t=6.0\,\text{southward}\enspace y=-32.four\,\text{thou}\enspace ten=108\,\text{g}[/latex]

Trying to escape his pursuers, a secret agent skis off a slope inclined at [latex]30^\circ[/latex] below the horizontal at sixty km/h. To survive and land on the snow 100 grand below, he must clear a gorge 60 m broad. Does he go far? Ignore air resistance.

A golfer on a fairway is 70 thousand away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of [latex]40^\circ[/latex] with an initial speed of twenty m/s, how close to the green does she come?

Testify Solution

[latex]{v}_{0y}=12.ix\,\text{m}\text{/}\text{s}\,y-{y}_{0}={v}_{0y}t-\frac{ane}{2}g{t}^{2}\enspace -20.0=12.9t-four.9{t}^{ii}[/latex]

[latex]t=three.7\,\text{s}\enspace{v}_{0x}=15.3\,\text{one thousand}\text{/}\text{s}\Rightarrow x=56.seven\,\text{1000}[/latex]

So the golfer's shot lands 13.3 thou short of the dark-green.

A projectile is shot at a hill, the base of which is 300 m abroad. The projectile is shot at [latex]60^\circ[/latex] above the horizontal with an initial speed of 75 yard/south. The hill can be approximated by a airplane sloped at [latex]20^\circ[/latex] to the horizontal. Relative to the coordinate system shown in the following figure, the equation of this direct line is [latex]y=(\text{tan}20^\circ)x-109.[/latex] Where on the loma does the projectile land?

An astronaut on Mars kicks a soccer ball at an angle of [latex]45^\circ[/latex] with an initial velocity of fifteen m/south. If the acceleration of gravity on Mars is three.seven m/s, (a) what is the range of the soccer kick on a flat surface? (b) What would exist the range of the same kick on the Moon, where gravity is one-sixth that of Earth?

Evidence Solution

a. [latex]R=threescore.8\,\text{m}[/latex],

b. [latex]R=137.8\,\text{grand}[/latex]

Mike Powell holds the record for the long jump of 8.95 thou, established in 1991. If he left the ground at an angle of [latex]fifteen^\circ,[/latex] what was his initial speed?

MIT'south robot cheetah can jump over obstacles 46 cm high and has speed of 12.0 km/h. (a) If the robot launches itself at an bending of [latex]60^\circ[/latex] at this speed, what is its maximum elevation? (b) What would the launch angle have to be to reach a height of 46 cm?

Evidence Solution

a. [latex]{five}_{y}^{two}={v}_{0y}^{2}-2gy\Rightarrow y=2.9\,\text{thou}\text{/}\text{s}[/latex]

[latex]y=iii.iii\,\text{1000}\text{/}\text{s}[/latex]

[latex]y=\frac{{v}_{0y}^{2}}{2g}=\frac{{({v}_{0}\text{sin}\,\theta )}^{ii}}{2g}\Rightarrow \text{sin}\,\theta =0.91\Rightarrow \theta =65.five^\circ[/latex]

Mt. Asama, Japan, is an agile volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the crater. If the volcanic rocks were launched at an bending of [latex]40^\circ[/latex] with respect to the horizontal and landed 900 m below the crater, (a) what would be their initial velocity and (b) what is their time of flight?

Drew Brees of the New Orleans Saints can throw a football game 23.0 chiliad/south (fifty mph). If he angles the throw at [latex]10^\circ[/latex] from the horizontal, what altitude does it get if information technology is to exist caught at the same peak as it was thrown?

Show Solution

[latex]R=eighteen.5\,\text{m}[/latex]

The Lunar Roving Vehicle used in NASA's tardily Apollo missions reached an unofficial lunar state speed of five.0 m/s by astronaut Eugene Cernan. If the rover was moving at this speed on a flat lunar surface and striking a small bump that projected it off the surface at an angle of [latex]20^\circ,[/latex] how long would information technology be "airborne" on the Moon?

A soccer goal is 2.44 m loftier. A histrion kicks the brawl at a distance 10 m from the goal at an angle of [latex]25^\circ.[/latex] What is the initial speed of the soccer ball?

Show Solution

[latex]y=(\text{tan}\,{\theta }_{0})x-[\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{ii}}]{10}^{2}\Rightarrow {five}_{0}=16.4\,\text{one thousand}\text{/}\text{s}[/latex]

Olympus Mons on Mars is the largest volcano in the solar system, at a elevation of 25 km and with a radius of 312 km. If you are continuing on the summit, with what initial velocity would you have to burn down a projectile from a cannon horizontally to clear the volcano and land on the surface of Mars? Notation that Mars has an dispatch of gravity of [latex]3.seven\,\text{m}\text{/}{\text{south}}^{2}.[/latex]

In 1999, Robbie Knievel was the first to bound the Grand Canyon on a motorcycle. At a narrow part of the canyon (69.0 m wide) and traveling 35.viii g/south off the takeoff ramp, he reached the other side. What was his launch angle?

Show Solution

[latex]R=\frac{{v}_{0}^{2}\text{sin}\,2{\theta }_{0}}{g}\Rightarrow {\theta }_{0}=15.0^\circ[/latex]

You throw a baseball game at an initial speed of 15.0 m/s at an angle of [latex]30^\circ[/latex] with respect to the horizontal. What would the ball's initial speed have to be at [latex]30^\circ[/latex] on a planet that has twice the acceleration of gravity as Earth to accomplish the same range? Consider launch and bear on on a horizontal surface.

Aaron Rogers throws a football game at 20.0 thou/due south to his wide receiver, who runs straight down the field at ix.iv thousand/s for 20.0 m. If Aaron throws the football when the wide receiver has reached 10.0 one thousand, what angle does Aaron have to launch the ball then the receiver catches it at the 20.0 one thousand mark?

Testify Solution

It takes the broad receiver 1.1 southward to cover the last 10 m of his run.

[latex]{T}_{\text{tof}}=\frac{2({v}_{0}\text{sin}\,\theta )}{thou}\Rightarrow \text{sin}\,\theta =0.27\Rightarrow \theta =15.6^\circ[/latex]

Glossary

projectile motion

motion of an object bailiwick only to the acceleration of gravity

range

maximum horizontal distance a projectile travels

time of flight

elapsed time a projectile is in the air

trajectory

path of a projectile through the air

A Dart Is Thrown Horizontally,

Source: https://pressbooks.online.ucf.edu/phy2048tjb/chapter/4-3-projectile-motion/

Posted by: flynncrue1941.blogspot.com

Share this post