You are throwing a baseball in the gym (because it is wet and cold outside). Aft

Question

You are throwing a baseball in the gym (because it is wet and cold outside). After warming up (and calibrating your arm) you are pretty sure you can throw a baseball with an initial velocity of 20 m/s. you figure your release point to be 2 meters above the floor and the ceiling of the gym is 8 meters high. In order to maximize the distance(horizontally) that you can throw the ball, what must be the initial angle of release? How far does the ball travel (in the air-not counting rolling) and what is its speed when it lands on the floor?

Explanation / Answer

Hmax = 8-2 = 6m

6 = u^2*Sin^2Q / 2g

Q = 32.85 degree

t1 = 20*Sin32.85 / g = 1.106 sec

v^2 = 2gH

Vy = 12.528 m/s

t2 = Vy / g = 1.277 s

Tnet = 2.383 sec

Range = 20*Cos32.85*2.383 = 47.68 m

V final = sqrt (Vy^2 + Ux^2)

V = 20.958 m/s

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