Question

A ball is thrown straight up from the ground with an initial velocity of 31.2 m/s; at the same instant, a ball is dropped from the roof of a building 21.3 m high. After how long will the balls be at the same height

Answer 1

Answer:

S1 = V1 t + 1/2 g t^2      distance traveled by ball dropped

S1 = 1/2 g t^2        since ball is dropped and V1 = zero

S2 = -1/2 g t^2 + V2 t        height to which ball 2 rises in time t

S1 + S2 = V2 t     both balls at same height after time t

t = S / V2 = 21.3 / 31.2 = .683 sec

Check:

S1 = 1/2 g t^2 = 9.8 / 2 * .683^2 = 2.29 m

S2 = 31.2 * .683 - 9.8/2 * .683^2 = 19 m

Total distance traveled = 2.29 + 19 = 21.3 m