5.5 projectile motion

In several earlier sections, we discussed aspects of the motion of an object moving back and forth along a straight line (rectilinear motion). You should recall that we determined that if we know the position of an object at any time t, then we can determine its velocity and acceleration, by differentiation. Specifically, the derivative of the position function is velocity and the derivative of velocity is acceleration. A much more important problem is to go backward, that is, to find the position and velocity of an object, given its acceleration. Mathematically, this means that, starting with the derivative of a function, we must find the original function. Now that we have integration at our disposal, we can accomplish this with ease. In this section, we also consider the case of an object moving along a curve in two dimensions, rather than the simpler case of rectilinear motion considered earlier.

You may already be familiar with Newton's second law of motion, which says that

F = ma,
where F is the sum of the forces acting on an object, m is the mass of the object and a is the acceleration of the object. This says that if you know all of the forces acting on an object, you can determine its acceleration. More importantly, once you know some calculus, you can determine an object's velocity and position from its acceleration.


Start by imagining that you take a dive from a high diving platform. The primary force acting on you throughout the dive is gravity. The force due to gravity is your own weight, which is related to mass by W = mg , where g is the gravitational constant. (Common approximations of g, accurate near sea level are 32 ft/s2 and 9.8 m/s2.) To keep the problem simple mathematically, we will ignore any other forces, such as air resistance, acting on you during your dive.

Let h(t ) represent your height above the water t seconds after jumping. Then the force due to gravity is F = -mg , where the minus sign indicates that the force is acting downward, in the negative direction. From our earlier work, the acceleration is a(t ) = h'' (t ). Newton's second law gives us -mg = mh'' (t ) or

h'' (t ) = -g.
Notice that the derivation of this equation depended only on the identification of gravity as the sole force of interest. The position function of any object (regardless of its mass) subject to gravity and no other forces will satisfy the same equation. The only differences from situation to situation are the initial conditions (the initial velocity and initial position) and the questions being asked.

5.1   
Finding the Velocity of a Diver at Impact
 
If a diving board is 15 feet above the surface of the water and a diver starts with initial velocity 8 ft/s (in the upward direction), what is the diver's velocity at impact (assuming no air resistance)?
 
 
If the height (in feet above the water) at time t is given by h(t ), Newton's second law gives us h'' (t ) = -32 (since the height is given in feet). Further, we have the initial conditions h(0) = 15 (the diver starts 15 feet above the water) and h' (0) = 8. As you'll see, finding h(t ) takes little more than elementary integration. We have
or
h' (t ) = -32t+c.
Notice that from the initial velocity, we have
8 = h' (0) = -32(0) +c = c,
so that the velocity at any time t is given by
h' (t ) = -32t+8.
To find the velocity at impact, you first need to find the time of impact. Notice that the diver will impact the water when h(t ) = 0 (i.e., when the height above the water is 0), so you need to perform one more integration to find the height function. Integrating the velocity function gives us
or
h(t ) = -16t 2+8t+c.
From the initial height, we have
15 = h(0) = -16(0) 2+8(0) +c = c,
so that
h(t ) = -16t 2+8t+15.
This gives the height above the water at any time t. Impact then occurs when
  0 = h(t ) = -16t 2+8t+15
  = -(4t+3) (4t-5),
so that t = is the time of impact. (You may ignore the extraneous solution t = -. Why?). When t = , the velocity is ft/s (impact velocity). To put this in more familiar units of velocity, multiply by 3600/5280 to convert to miles per hour. In this case, the impact velocity is about 22 mph. (You probably don't want to come down in the wrong position at that speed!)
 

In example 5.1, the negative sign of the velocity indicated that the diver was coming down. In many situations, both upward and downward motions are important.



 

 

 
5.2   
An Equation for the Vertical Motion of a Ball
 
A ball is propelled from the ground straight upward with initial velocity 64 ft/s. Ignoring air resistance, find an equation for the height of the ball at any time t. Also, determine the maximum height and the amount of time the ball spends in the air.
 
 
With gravity as the only force, the height h(t ) satisfies h'' (t ) = -32. The initial conditions are h' (0) = 64 and h(0) = 0. We then have
or
h' (t ) = -32t+c.
From the initial velocity, we have
64 = h' (0) = -32(0) +c = c
and so,
h' (t ) = 64-32t.
Integrating one more time gives us
or
h(t ) = 64t-16t 2+c.
From the initial altitude we have
0 = h(0) = 64(0) -16(0) 2+c = c,
and so,
h(t ) = 64t-16t 2.

Since the height function is quadratic, its maximum occurs at the one time when h' (t ) = 0. [You should also consider the physics of the situation: what happens physically when h' (t ) = 0 ?] Solving 64-32t = 0 gives t = 2 (the time at the maximum altitude) and the corresponding height is h(2) = 64(2)-16(2)2 = 64 feet. Again, the ball lands when h(t ) = 0. Solving

0 = h(t ) = 64t-16t 2 = 16t (4-t )
gives t = 0 (launch time) and t = 4 (landing time). The time of flight is thus 4 seconds.
 

You can observe an interesting property of projectile motion by graphing the height function from example 5.2 along with the line y = 48 (see Figure 5.43).



Figure 5.43
Height of the ball at time t.

Notice that the graphs intersect at t = 1 and t = 3. Further, the time interval [1, 3] corresponds to exactly half the total time in the air. Notice that this says that the ball is in the top one-fourth of its height for half of its time in the air. You may have marveled at how some athletes jump so high that they seem to “hang in the air. ” As this calculation suggests, all objects tend to hang in the air.

5.3   
Finding the Initial Velocity Required to Reach a Certain Height
 
It has been reported that basketball star Michael Jordan has a vertical leap of 54''. Ignoring air resistance, what is the initial velocity required to jump this high?
 
 
Once again, Newton's second law leads us to the equation h'' (t ) = -32 for the height h(t ). Further, the initial velocity and initial position are given by h' (0) = v0 and h(0) = 0, respectively. Our task is to determine the value of v0 that will give a maximum altitude of 54''. As before, we integrate to get
h' (t ) = -32t+c.
Using the initial velocity, we get
v0 = h' (0) = -32(0) +c = c.
This gives us the velocity function
h' (t ) = v0-32t.
Integrating once again and using the initial position h(0) = 0, we get
h(t ) = v0t-16t 2.
The maximum height occurs when h' (t ) = 0. (Why?) Setting
0 = h' (t ) = v0-32t,
gives us . The height at this time (i.e., the maximum altitude) is then
A jump of 54'' = 4.5' requires or v02 = 288, so that v0 = 17 ft/s. Note that this initial velocity is equivalent to roughly 11.6 mph.
 

So far, our projectiles have only traveled in the vertical direction. Most applications of projectile motion must also consider movement in the horizontal direction. If we ignore air resistance, these calculations are also relatively straightforward. The basic principle is that we can apply Newton's second law separately for the horizontal and vertical components of the motion. If y(t ) represents the vertical position, then we have y'' (t ) = -g, as before. Ignoring air resistance, there are no forces acting horizontally on the projectile. So, if x(t ) represents the horizontal position, Newton's second law gives us x'' (t ) = 0.

The initial conditions are slightly more complicated here. In general, we want to consider projectiles that are launched with an initial speed v0 at an angle from the horizontal. Figure 5.44a shows a projectile fired with > 0.



Figure 5.44a
Path of projectile.

Notice that an initial angle of < 0 would mean a downward initial velocity.

As shown in Figure 5.44b, the initial velocity can be separated into horizontal and vertical components. From elementary trigonometry, the horizontal component of the initial velocity is vx = v0cos and the vertical component is vy = v0sin .



Figure 5.44b
Vertical and horizontal components of velocity.

5.4   
The Motion of a Projectile in Two Dimensions
 
An object is launched at angle = /6 from the horizontal with initial speed v0 = m/s. Determine the time of flight and the (horizontal) range of the projectile.
 
 
Starting with the vertical component of the motion (and again ignoring air resistance), we have y'' (t ) = -9.8 (since the initial speed is given in terms of meters per second). Referring to Figure 5.44b, notice that the vertical component of the initial velocity is y' (0) = 98sin /6 = 49 and the initial altitude is y(0) = 0. A pair of simple integrations give us the velocity function y' (t ) = -9.8t+49 and the position function y(t ) = -4.9t 2+49t . The ball hits the ground when y(t ) = 0 (i.e., when its height above the ground is 0). Solving
0 = y(t ) = -4.9t 2+49t = 49t (1-0.1t )
gives t = 0 (launch time) and t = 10 (landing time). The time of flight is then 10 seconds. The horizontal component of motion is determined from the equation x'' (t ) = 0 with initial velocity x' (0) = 98cos /6 = 49 and initial position x(0) = 0. Integration gives us x' (t ) = 49 and x(t ) = (49)t . In Figure 5.45, we give a plot of the path of the ball. [You can easily do the same by using the parametric plot mode on your graphing calculator or CAS. Simply enter the separate equations for x(t ) and y(t ) and set the range of t - values to be 0 t 10. Alternatively, you could easily solve for t in terms of x and replace t by to see that the curve is simply a parabola.]



Figure 5.45
Path of ball.

Notice that the horizontal location of the ball when it lands is the value of x(t ) at t = 10 (the landing time). The range is then
x(10) = (49)(10) = 490 849 meters.

5.1
5.5   
The Motion of a Tennis Serve
 
Suppose that a tennis player hits a serve from a height of 8 feet at an initial speed of 120 mph and at an angle of 5° below the horizontal. The serve is “in” if the ball clears a 3 ' - high net that is 39 ' away and hits the ground in front of the service line 60' away. (We illustrate this situation in Figure 5.46.) Determine whether the serve is in or out.
 
 
As in example 5.4, we start with the vertical motion of the ball. Since distance is given in feet, the equation of motion is y''(t ) = -32. The initial speed must be converted to feet per second: 120 mph = 120 ft/s = 176 ft/s. The vertical component of the initial velocity is then y' (0) = 176sin(-5°) -15.34 ft/s. Integration then gives us
y'(t ) = -32t-15.34.



Figure 5.46
Height of tennis serve.

The initial height is y(0) = 8 ft, so another integration gives us
y(t ) = -16t 2-15.34t+8 ft.
The horizontal component of motion is determined from x''(t ) = 0 with initial velocity x'(0) = 176 cos(-5°) 175.33 ft/s and initial position x(0) = 0. Integrations give us x'(t ) = 175.33 ft/s and x(t ) = 175.33t  ft. Summarizing, we have
  x(t ) = 175.33t, y(t ) = -16t 2-15.34t+8.
For the ball to clear the net, y must be at least 3 when x = 39. We have x(t ) = 39 when 175.33t = 39 or t 0.2224. At this time, y(0.2224) 3.8, showing that the ball is high enough to clear the net. The second requirement is that we need to have x 60 when the ball lands (y = 0). We have y(t ) = 0 when -16t 2-15.34t+8 = 0. From the quadratic formula, we get t -1.33 and t 0.3749. Ignoring the negative solution, we compute x(0.3749) 65.7, so that the serve lands beyond the service line. The serve is not in.
 

One reason to start with Newton's second law is to force yourself to consider the forces that are (and are not) being considered. For example, we have thus far ignored air resistance. It is important to realize that this is a simplification of reality. Some calculations using such simplified equations are reasonably accurate. Others, such as in the following example, are not.

5.6   
An Example Where Air Resistance Can't be Ignored
 
Suppose a rain drop falls from a cloud 3000 feet above the ground. Ignoring air resistance, how fast would the raindrop be falling when it hits the ground?
 
 
If the height of the raindrop at time t is given by y(t ), Newton's second law of motion tells us that y'' (t ) = -32. Further, we have the initial velocity y' (0) = 0 (since the drop falls as opposed to being thrown down) and the initial altitude y(0) =. Integrating and using the initial conditions gives us y' (t ) = -32t  and y(t ) = 3000-16t 2. The raindrop hits the ground when y(t ) = 0. Setting
0 = y(t ) = 3000-16t 2
gives us seconds. The velocity at this time is then
This corresponds to nearly 300 mph! Fortunately, air resistance does play a significant role in the fall of a raindrop, which has an actual landing speed of about 10 mph.
 

The obvious lesson from example 5.6 is that it is not always reasonable to ignore air resistance. Some of the mathematical tools needed to more fully analyze projectile motion with air resistance are developed in Chapter 6.

The air resistance (more precisely, air drag) that slows the raindrop down is only one of the ways in which air can affect the motion of an object. The Magnus force, produced by the spinning of an object or asymmetries (i.e., lack of symmetry) in the shape of an object, can cause the object to change directions and curve. Perhaps the most common example of a Magnus force occurs on an airplane. One side of an airplane wing is curved and the other side is comparatively flat (see Figure 5.47).



Figure 5.47
Cross section of a wing.

The lack of symmetry causes the air to move over the top of the wing faster than it moves over the bottom. This produces a Magnus force in the upward direction (lift), lifting the airplane into the air.

A more down-to-earth example of a Magnus force occurs in an unusual baseball pitch called the knuckleball. To throw this pitch, the pitcher grips the ball with all five fingers and tries to throw the ball with as little spin as possible (despite the name, the knuckles are not used). Baseball players claim that the knuckleball “dances” around unpredictably and is exceptionally hard to hit or catch. There still is no complete agreement on why the knuckleball moves so much, but we will present one current theory due to physicists Robert Watts and Terry Bahill.

The cover of the baseball is sewed on with stitches that are raised up slightly from the rest of the ball.



Regulation baseball, showing stitching.

These curved stitches act much like an airplane wing, creating a Magnus force that affects the ball. The direction of the Magnus force depends on the exact orientation of the ball's stitches. Measurements by Watts and Bahill indicate that the lateral force (left/right from the pitcher's perspective) is approximately Fm = -0.1sin (4) lb, where is the angle in radians of the ball's position rotated from a particular starting position.

Since gravity does not affect the lateral motion of the ball, the only force acting on the ball laterally is the Magnus force. Newton's second law applied to the lateral motion of the knuckleball gives mx'' (t ) = -0.1sin (4). The mass of a baseball is about 0.01 slugs. (If you are unfamiliar with slugs, these are the units of measurement of mass in the English system. To get the more familiar weight in pounds, simply multiply the mass by g = 32. )

x'' (t ) = -10sin (4).
If the ball is spinning at the rate of radians per second, then 4 = 4 t+ 0 , where the initial angle 0 depends on where the pitcher grips the ball. Our model is then
x'' (t ) = -10sin (4 t+ 0), (5.1)
with initial conditions x' (0) = 0 and x(0) = 0. For a typical knuckleball speed of 60 mph, it takes about 0.68 second for the pitch to reach home plate.

5.7   
An Equation for the Motion of a Knuckleball
 
For a spin rate of = 2 radians per second and 0 = 0 , find an equation for the motion of the knuckleball and graph it for 0 t 0.68. Repeat this for 0 = /2.
 
 
For 0 = 0 , Newton's second law gives us x'' (t ) = -10sin 8t  , from (5.1). Integrating this and using the initial condition x' (0) = 0 gives us
x' (t ) = -[- cos 8t-(- cos 0) ] = 1.25(cos 8t-1).
Integrating once again and using the second initial condition x(0) = 0 , we get
A graph of this function shows the lateral motion of the ball (see Figure 5.48a).



Figure 5.48a
Lateral motion of a knuckleball for 0 = 0.

The graph shows the path of the pitch as it would look viewed from above. Notice that after starting out straight, this pitch breaks nearly a foot away from the center of home plate! For the case where 0 = /2 , we have from (5.1) that
Integrating this and using the first initial condition gives us
Integrating a second time yields
A graph of the lateral motion in this case is shown in Figure 5.48b.



Figure 5.48b
Lateral motion of a knuckleball for

Notice that this pitch breaks nearly 4 inches to the pitcher's right and then curves back over the plate for a strike! You can see that in theory, the knuckleball is very sensitive to spin and initial position, and can be very difficult to hit when thrown properly.
 

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