Projectile Motion Example Problem 3

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 v0 and q.

3. This v = f(q) function defines the set (an infinite number) of all combinations of v0 and q for trajectories that will hit the target.

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

9. 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).

10. 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

11. 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 and a higher arc.

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