1. Projectile Motion Example Problem 1 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, v 0 ; (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.

2. First establish an x-y coordinate system that makes sense for the problem:

3. Write the two projectile position equations: x = x 0 + (v 0 ·cos  · ty = y 0 + (v 0 · sin  ) · t – ½gt 2 15 = 0 + (v 0 ·cos 50  · t 10 = 7 + (v 0 · sin 50) · t – ½(32.2)t 2 15 =.643 · v 0 · t 3 =.766 · v 0 · t – 16.1 · t 2 It’s easy to solve these two equations (see the next page) for the two unknowns (v 0 and t).

4. There are several ways to solve for v 0 and t, but I prefer to substitute the v 0 t product from the first equation into the second: 15 =.643 · v 0 · t 3 =.766 · v 0 · t – 16.1 · t 2 23.34 = v 0 · t3 =.766 · (23.34)– 16.1 · t 2 16.1 · t 2 = 17.88 – 3 = 14.88 t = 0.961 sec 23.34 = v 0 · t v 0 = 24.3 fps

5. Now find the height of the apex of the trajectory: v 0 = 24.3 fpsEasiest equation to use for this: v y 2 = v 0y 2 – 2g(y – y 0 ) v 0y = 24.3(sin 50) = 18.6 fps 0 = (18.6) 2 – 2(32.2)(y - 7) y = h max = 12.38 ft This is a low trajectory. A good free throw should have a larger launch angle and a higher arc.

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