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Chapter 13: Vector-Valued Functions and Motion in Space
13.2
Modeling Projectile Motion
When we shoot a projectile into the air we usually want to know beforehand how far it will go (will it reach the target?), how high it will rise (will it clear the hill?), and when it will land (when do we get results?). We get this information from the direction and magnitude of the projectiles initial velocity vector, using Newtons second law of motion.
The Vector and Parametric Equations for Ideal Projectile Motion
To derive equations for projectile motion, we assume that the projectile behaves like a particle moving in a vertical coordinate plane and that the only force acting on the projectile during its flight is the constant force of gravity, which always points straight down. In practice, none of these assumptions really holds. The ground moves beneath the projectile as the earth turns, the air creates a frictional force that varies with the projectiles speed and altitude, and the force of gravity changes as the projectile moves along. All this must be taken into account by applying corrections to the predictions of the ideal equations we are about to derive. The corrections, however, are not the subject of this section. We assume that the projectile is launched from the origin at time t = 0 into the first quadrant with an initial velocity v0 (Figure 13.9). If v0 makes an angle a with the horizontal, then v0 = s v0 cos a di + s v0 sin a dj. If we use the simpler notation y0 for the initial speed v0 , then v0 = s y0 cos a di + s y0 sin a dj. The projectiles initial position is r0 = 0i + 0j = 0.
(2) (1)
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Newtons second law of motion says that the force acting on the projectile is equal to the projectiles mass m times its acceleration, or ms d 2r> dt 2 d if r is the projectiles position vector and t is time. If the force is solely the gravitational force - mg j, then m
v0 v0 sin j x
d 2r = - mg j dt 2
and
d 2r = - g j. dt 2
We find r as a function of t by solving the following initial value problem. Differential equation: Initial conditions: The first integration gives d 2r = -gj dt 2 r = r0 and dr = v0 dt when t = 0
r 0 at time t 0
v0 cos i a g j (a)
y (x, y) v g j
dr = - s gt dj + v0 . dt A second integration gives r=12 gt j + v0 t + r0 . 2
a r 0 xi yj
Substituting the values of v0 and r0 from Equations (1) and (2) gives
x R Horizontal range (b)
r=-
12 gt j + s y0 cos a dt i + s y0 sin a dt j + 0 2 ('''''')''''''*
v0 t
FIGURE 13.9 (a) Position, velocity, acceleration, and launch angle at t = 0 . (b) Position, velocity, and acceleration at a later time t.
Collecting terms, we have
Ideal Projectile Motion Equation r = s y0 cos a dt i + as y0 sin a dt 12 gt b j. 2
(3)
Equation (3) is the vector equation for ideal projectile motion. The angle a is the projectiles launch angle (firing angle, angle of elevation), and y0 , as we said before, is the projectiles initial speed. The components of r give the parametric equations x = s y0 cos a dt and y = s y0 sin a dt 12 gt , 2
(4)
where x is the distance downrange and y is the height of the projectile at time t 0.
EXAMPLE 1
Firing an Ideal Projectile
A projectile is fired from the origin over horizontal ground at an initial speed of 500 m> sec and a launch angle of 60. Where will the projectile be 10 sec later?
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Chapter 13: Vector-Valued Functions and Motion in Space Solution
We use Equation (3) with y0 = 500, a = 60, g = 9.8, and t = 10 to find the projectiles components 10 sec after firing. r = s y0 cos a dt i + as y0 sin a dt 23 1 1 = s 500 d a b s 10 di + as 500 d a b 10 - a b s 9.8 ds 100 d b j 2 2 2 12 gt b j 2
L 2500i + 3840j. Ten seconds after firing, the projectile is about 3840 m in the air and 2500 m downrange.
Height, Flight Time, and Range
Equation (3) enables us to answer most questions about the ideal motion for a projectile fired from the origin. The projectile reaches its highest point when its vertical velocity component is zero, that is, when dy = y0 sin a - gt = 0, dt For this value of t, the value of y is ymax = s y0 sin a d a s y0 sin a d2 y0 sin a y0 sin a 2 1 b - ga g b = . g 2 2g or t= y0 sin a g.
To find when the projectile lands when fired over horizontal ground, we set the vertical component equal to zero in Equation (3) and solve for t. s y0 sin a dt t ay0 sin a 12 gt = 0 2 1 gt b = 0 2 t = 0, t= 2y0 sin a g
Since 0 is the time the projectile is fired, s 2y0 sin a d> g must be the time when the projectile strikes the ground. To find the projectiles range R, the distance from the origin to the point of impact on horizontal ground, we find the value of the horizontal component when t = s 2y0 sin a d> g . x = s y0 cos a dt R = s y0 cos a d a 2y0 sin a y02 y02 b = g s 2 sin a cos a d = g sin 2a g
The range is largest when sin 2a = 1 or a = 45.
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Height, Flight Time, and Range for Ideal Projectile Motion For ideal projectile motion when an object is launched from the origin over a horizontal surface with initial speed y0 and launch angle a : Maximum height: Flight time: Range: ymax = t= s y0 sin a d2 2g 2y0 sin a g
y02 R = g sin 2a .
EXAMPLE 2
Investigating Ideal Projectile Motion
Find the maximum height, flight time, and range of a projectile fired from the origin over horizontal ground at an initial speed of 500 m> sec and a launch angle of 60 (same projectile as Example 1).
Solution
Maximum height: ymax = = Flight time: t
s y0 sin a d2 2g
s 500 sin 60 d2 L 9566 m 2s 9.8 d 2y0 sin a = g = 2s 500 d sin 60 L 88.4 sec 9.8
Range:
y 10,000 8000 6000 4000 2000 0 2000 5000 10,000 15,000 20,000 x
R
y02 = g sin 2a = s 500 d2 sin 120 L 22,092 m 9.8
From Equation (3), the position vector of the projectile is r = s y0 cos a dt i + as y0 sin a dt 12 gt b j 2 1 s 9.8 dt 2 b j 2
= s 500 cos 60 dt i + as 500 sin 60 dt = 250t i +
FIGURE 13.10 The graph of the projectile described in Example 2.
A A 250 23 B t - 4.9t 2 B j.
A graph of the projectiles path is shown in Figure 13.10.
Ideal Trajectories Are Parabolic
It is often claimed that water from a hose traces a parabola in the air, but anyone who closely looks enough will see this is not so. The air slows the water down, and its forward progress is too slow at the end to keep pace with the rate at which it falls.
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Chapter 13: Vector-Valued Functions and Motion in Space
What is really being claimed is that ideal projectiles move along parabolas, and this we can see from Equations (4). If we substitute t = x> s y0 cos a d from the first equation into the second, we obtain the Cartesian-coordinate equation
v0
y = -a
g 2y02 cos2 a
b x 2 + s tan a dx .
(x 0 , y0 ) x
This equation has the form y = ax 2 + bx , so its graph is a parabola.
0
Firing from s x0, y0 d
If we fire our ideal projectile from the point s x0, y0 d instead of the origin (Figure 13.11), the position vector for the path of motion is r = s x0 + s y0 cos a dt di + ay0 + s y0 sin a dt as you are asked to show in Exercise 19. 12 gt b j, 2
(5)
FIGURE 13.11 The path of a projectile fired from s x0 , y0 d with an initial velocity v0 at an angle of a degrees with the horizontal.
EXAMPLE 3
Firing a Flaming Arrow
To open the 1992 Summer Olympics in Barcelona, bronze medalist archer Antonio Rebollo lit the Olympic torch with a flaming arrow (Figure 13.12). Suppose that Rebollo shot the arrow at a height of 6 ft above ground level 90 ft from the 70-ft-high cauldron, and he wanted the arrow to reach maximum height exactly 4 ft above the center of the cauldron (Figure 13.12).
Photograph is not available.
FIGURE 13.12 Spanish archer Antonio Rebollo lights the Olympic torch in Barcelona with a flaming arrow.
(a) Express ymax in terms of the initial speed y0 and firing angle a . (b) Use ymax = 74 ft (Figure 13.13) and the result from part (a) to find the value of y0 sin a . (c) Find the value of y0 cos a . (d) Find the initial firing angle of the arrow.
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ymax
74'
(a) We use a coordinate system in which the positive x-axis lies along the ground toward the left (to match the second photograph in Figure 13.12) and the coordinates of the flaming arrow at t = 0 are x0 = 0 and y0 = 6 (Figure 13.13). We have
6'
x
90'
NOT TO SCALE
0
y = y0 + s y0 sin a dt = 6 + s y0 sin a dt -
12 gt 2 12 gt . 2
Equation (5), j-component
y0 = 6
FIGURE 13.13 Ideal path of the arrow that lit the Olympic torch (Example 3).
We find the time when the arrow reaches its highest point by setting dy> dt = 0 and solving for t, obtaining t= For this value of t, the value of y is ymax = 6 + s y0 sin a d a =6+ y0 sin a y0 sin a 2 1 g b - 2 ga g b y0 sin a g.
s y0 sin a d2 . 2g
(b) Using ymax = 74 and g = 32, we see from the preceeding equation in part (a) that y0 sin a = 2s 68 ds 64 d . 74 = 6 + s y0 sin a d2 2s 32 d
or
(c) When the arrow reaches ymax , the horizontal distance traveled to the center of the cauldron is x = 90 ft. We substitute the time to reach ymax from part (a) and the horizontal distance x = 90 ft into the i-component of Equation (5) to obtain x = x0 + s y0 cos a dt 90 = 0 + s y0 cos a dt = s y0 cos a d a y0 sin a g b.
Equation (5), i-component x = 90, x0 = 0 t = s y0 sin a d> g
Solving this equation for y0 cos a and using g = 32 and the result from part (b), we have y0 cos a = (d) Parts (b) and (c) together tell us that y0 sin a tan a = y0 cos a =
90g s 90 ds 32 d = . y0 sin a 2s 68 ds 64 d
A 2s 68 ds 64 d B 2
s 90 ds 32 d
=
68 45
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or a = tan-1 a This is Rebollos firing angle. 68 b L 56.5. 45
Projectile Motion with Wind Gusts
The next example shows how to account for another force acting on a projectile. We also assume that the path of the baseball in Example 4 lies in a vertical plane.
EXAMPLE 4
Hitting a Baseball
A baseball is hit when it is 3 ft above the ground. It leaves the bat with initial speed of 152 ft> sec, making an angle of 20 with the horizontal. At the instant the ball is hit, an instantaneous gust of wind blows in the horizontal direction directly opposite the direction the ball is taking toward the outfield, adding a component of - 8.8i s ft> sec d to the balls initial velocity s 8.8 ft> sec = 6 mph d . (a) Find a vector equation (position vector) for the path of the baseball. (b) How high does the baseball go, and when does it reach maximum height? (c) Assuming that the ball is not caught, find its range and flight time.
Solution
(a) Using Equation (1) and accounting for the gust of wind, the initial velocity of the baseball is v0 = s y0 cos a di + s y0 sin a dj - 8.8i = s 152 cos 20 di + s 152 sin 20 dj - s 8.8 di = s 152 cos 20 - 8.8 di + s 152 sin 20 dj. The initial position is r0 = 0i + 3j. Integration of d 2r> dt 2 = - g j gives dr = - s gt dj + v0 . dt A second integration gives r=12 gt j + v0 t + r0 . 2
Substituting the values of v0 and r0 into the last equation gives the position vector of the baseball. r=12 gt j + v0 t + r0 2
= s 152 cos 20 - 8.8 dt i + A 3 + (152 sin 20 dt - 16t 2 B j.
= - 16t 2j + s 152 cos 20 - 8.8 dt i + s 152 sin 20 dt j + 3j
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(b) The baseball reaches its highest point when the vertical component of velocity is zero, or dy = 152 sin 20 - 32t = 0. dt Solving for t we find t= 152 sin 20 L 1.62 sec. 32
Substituting this time into the vertical component for r gives the maximum height ymax = 3 + s 152 sin 20 ds 1.62 d - 16s 1.62 d2 L 45.2 ft. That is, the maximum height of the baseball is about 45.2 ft, reached about 1.6 sec after leaving the bat. (c) To find when the baseball lands, we set the vertical component for r equal to 0 and solve for t: 3 + s 152 sin 20 dt - 16t 2 = 0 3 + s 51.99 dt - 16t 2 = 0. The solution values are about t = 3.3 sec and t = - 0.06 sec. Substituting the positive time into the horizontal component for r, we find the range R = s 152 cos 20 - 8.8 ds 3.3 d L 442 ft. Thus, the horizontal range is about 442 ft, and the flight time is about 3.3 sec. In Exercises 29 through 31, we consider projectile motion when there is air resistance slowing down the flight.
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