1. No category

Dr. Neal, WKU
MATH 117
Projectile Motion
Suppose we launch an aerodynamic object from initial height h0 ft, with initial velocity
v0 ft/s, at angle ! measured from the horizontal. We assume very little air resistance.
h0
!
The initial velocity is broken down into two components, the horizontal or x component, and the upward or y -component. Then
x = v0 cos !
and
y = v0 sin ! .
The object moves horizontally with constant x –direction velocity v0 cos ! until it
hits the ground. To find the length traveled at time t , we use (v0 cos ! ) " t ft.
The object initially moves upward with y –direction velocity v0 sin ! . However,
gravity diminishes this velocity by 32 ft/s every second. So the y –direction velocity is
v0 sin ! " 32t . The main results we need are:
The length and horizontal velocity are given by
length:
l(t) = (v0 cos ! )t ft
horizontal velocity: v x (t ) = v0 cos ! ft / s
The height and vertical velocity functions are then approximated by
height:
h(t) = !16t 2 + (v 0 sin " )t + h0
vertical velocity:
ft
v y (t) = !32 t + v0 sin " ft / s ,
(where gravity is g ≈ 32 ft / s2 ).
2
2
The overall speed at any time is given by s (t ) = ( v x (t) ) + ( v y (t) ) .
To find the time T needed to reach the maximum height, solve v y (t) = 0. Substitute
this time into h(t) to find the maximum height M . To find the time needed to hit the
ground, solve h(t) = 0. Substitute this time into l(t) to find the length traveled.
Note: The time needed to hit the ground is also given by T + M / 4 .
Dr. Neal, WKU
Example. Toss a solid object at 36 mph at a 35º angle from 4 ft off the ground. (a) How
high will it go? (b) Find the time needed to hit the ground. (c) How far will it go? (c)
With what speed will it hit the ground?
Solution. We first put everything into feet and ft/s:
Initial velocity is v0 = 36
mile
ft
1 hr
! 5280
!
= 52.8 ft/s.
hr
mile 3600 s
The necessary functions are then
height:
vertical velocity:
length:
horizontal velocity:
h(t) = !16 t 2 + (52.8 sin35º)t + 4 ≈ !16 t 2 + 30. 285 t + 4 ft
v y (t) = !32t + 30.285 ft/s
l(t ) = (52.8 cos35º)t ≈ 43.251 t ft
v x (t ) = 43.251 ft/s
(a) The maximum height occurs when the vertical velocity equals 0. So we solve
! 32 t + 30.285 = 0 , to obtain T ≈ 0.9464 sec. The maximum height is then
M = h(0.9464) ≈ !16(.9464)2 + 30.285(.9464) + 4 ≈ 18.33 ft.
(b) To find the time needed to hit the ground, we can use the quadratic formula (or
calculator) to solve h(t) = 0. That is, solve !16t 2 + 30.285 t + 4 = 0. The positive
solution will be
t=
!30.285 ! 30.2852 ! 4(!16)(4)
≈ 2.01677 sec
!32
But it is much easier to use T + M / 4 . So the time to hit the ground is
T+
M
18.33
! 0.9464 +
! 2.0167 sec
4
4
Note: There will be some slight error here due to the round-offs used in Part (a).
(c) To find horizontal distance, use the length function: 43. 251 ! 2.0167 ≈ 87.2 ft.
(d) To find the impact speed, we need the final components of velocity at time 2.0167
seconds. These components are v y (2. 0167) = !32 " 2. 01677 + 30.285 # !34.25 and
v x (2. 0167) = 43.251 .
Thus, the impact speed is
43. 2512 + 34.252 ≈ 55.17 ft/s ≈ 37.6 mph.
Dr. Neal, WKU
Exercises
1. Launch a solid object at 66 mph at a 46º angle from 6 ft off the ground. (a) How high
will it go? (b) Find the time needed to hit the ground. (c) How far will it go? (c) With
what speed will it hit the ground?
2. A pitcher throws a fastball at 90 mph from 5.8 feet off the ground in a direct
horizontal line. Find the time needed to hit the ground and how far the ball should go.
Challenge Problems
1. A mortar cannon launches from 1 foot off the ground with a muzzle velocity of 300
ft/s. Its target is 1200 feet away. Using a 45º angle, which maximizes the distance, does
it hit the target? What angle do we need to hit the target?
2. Peyton Manning throws a ball from 6 ft off the ground. Marvin Harrison catches it
3.2 seconds later, 4 ft off the ground and 50 yards downfield. With what initial velocity
and at what angle was the ball thrown?
Dr. Neal, WKU
Solutions to Exercises
mile
ft
1 hr
! 5280
!
1. Put everything in feet and ft/s: v0 = 66
= 96.8 ft/s. The
hr
mile 3600 s
necessary functions are then
height:
vertical velocity:
length:
horizontal velocity:
h(t) = !16 t 2 + (96.8 sin 46º )t + 6 ≈ !16 t 2 + 69.632 t + 6
v y (t) = !32t + 69.632
l(t ) = (96.8 cos46º )t ≈ 67.243 t
v x (t ) = 67. 243
(a) First solve !32t + 69.632 = 0, to obtain T ≈ 2.176 sec. The maximum height is then
M = h(2.176) ≈ 81.76 ft.
M
81.76
(b) The time to hit the ground is T +
! 2.176 +
! 4. 4365 sec
4
4
(c) The total horizontal length traveled is 67.243 ! 4.4365 ≈ 298.3 ft
(d)
The final components of velocity at time 4.4365
v x (4.4365) = 67. 243 and v y (4. 4365) = !32 " 4.4365 + 69. 632 = !72.336 .
seconds
are
2
2
Thus, the impact speed is 67.243 + 72. 336 ≈ 98.763 ft/s ≈ 67.338 mph.
2. Here ! = 0º , so sin ! = 0 and cos ! = 1 and all of the velocity is in the x -direction.
The height function is h(t) = !16 t 2 + 5.8 . If we solve h(t) = 0 , we obtain t = 5.8 / 4 ≈
0.602 seconds needed to hit the ground. (This time is also T + M / 4 = 0 + 5.8 / 4 .)
The initial velocity is 90 ! 5280 / 3600 = 132 ft/sec, all in the x -direction. So the ball
should travel 132 ! 0.602 ≈ 79.464 ft.