1. Projectile Motion Example Problem 3 In the previous problem an iron worker tossed a rivet at 10 m/s and we saw that two trajectories are possible for the rivet to strike the rim of the bucket on the scaffold. What if the worker tosses the rivet at a lower velocity? Is there a single minimum velocity and associated angle that will allow the rivet to just reach the rim of the bucket?

2. Where are we going? We’re seeking an equation relating v 0 and .

3. This v = f(  ) function defines the set (an infinite number) of all combinations of v 0 and  for trajectories that will hit the target.

4. Now that we have the v = f(  ) function, take the derivative with respect to , set it equal to zero, and find the minimum v and the associated angle.

9. 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).

10. 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

11. 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 and a higher arc.

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