Projectile Motion Example Problem 1

50° 10 ft

7 ft

15 ft

A player shoots a free throw to a basket 15 feet away and 10 feet off the floor. If the ball is released from a point 7 feet above the floor and at an angle of 50º, determine: (1) The required initial velocity, v0; (2) The time the ball passes through the rim; (3) The maximum height of the trajectory;(4) The speed of the ball and the angle of its trajectory as it passes through the rim.

First establish an x-y coordinate system that makes sense for theproblem:

50° 10 ft

7 ft

15 ft

y

x

15 ft

7 ft

10 ft

y

x

v = ?0

50°

Write the two projectile position equations:

x = x0 + (v0·cos ·t y = y0 + (v0·sin )·t – ½gt2

15 ft

7 ft

10 ft

y

x

v = ?0

50°

15 = 0 + (v0·cos 50·t 10 = 7 + (v0·sin 50)·t – ½(32.2)t2

15 = .643·v0·t 3 = .766·v0·t – 16.1·t2

It’s easy to solve these two equations (see the next page) for the two unknowns

(v0 and t).

There are several ways to solve for v0 and t, but I prefer to substitute the v0t product from the first equation into the second:

15 = .643·v0·t 3 = .766·v0·t – 16.1·t2

23.34 = v0·t 3 = .766·(23.34)– 16.1·t2

16.1·t2 = 17.88 – 3 = 14.88

t = 0.961 sec

23.34 = v0·t v0 = 24.3 fps

Now find the height of the apex of the trajectory:

v0 = 24.3 fps Easiest equation to use for this:

vy2 = v0y

2 – 2g(y – y0)v0y = 24.3(sin 50) = 18.6 fps

0 = (18.6)2 – 2(32.2)(y - 7)

y = hmax = 12.38 ft

This is a low trajectory. A good free throw should have a larger launch angle and a higher arc.

15 ft

7 ft

10 ft

y

x

v = ?0

50°

h = ?max

v = 0 at apexy

v 0y

v 0x

(x , y )= (0,7) ft

0 0

Finally, find the speed and angle of the ball as it passes through the rim:

v = 24.3 c os 50 = 15.62 fpsx

v = v s in 50 - gty 0

= 24.3 s in 50 - 32.2(.961)

= - 12.33 fps = 12.33 fps

vv = 15.6 + 12.32 2

= 19.9 fps

= tan -1 12.33

19.9= 38.3°