Question

A stone is dropped from a height of 49m and simultaneously another ball is thrown upward from the ground with a speed of 40m/s. when and where do the two stone meet
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Answer 1

Answer:

S1 = 1/2 g t^2     distance stone falls in time t

S2 = Vy t - 1/2 g t^2   distance thrown stone rises in time t

H = 49 = S1 + S2 = Vy t

t = 49 / 40 sec   time when stones meet

Check:

Stone 1 falls:      1/2 g t^2 = 1/2 * 9.8 * (49 / 40)^2 = 7.35 m

Stone 2 rises :  40 * (49 / 40) - 1/2 * 9.8 (49 / 40)^2 = 41.65 m