Projectile Motion Example Problem 1

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, 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.

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

3. Write the two projectile position equations: x = x0 + (v0·cos q)·t y = y0 + (v0·sin q)·t – ½gt2 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).

4. 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 v0 = 24.3 fps 23.34 = v0·t

5. Now find the height of the apex of the trajectory: v0 = 24.3 fps Easiest equation to use for this: vy2 = v0y2 – 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.

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