Question

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 ms^-1. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 ms^-1, and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hand?

Answer 1

When lift is stationary, u = 49 ms^-1, v = 0 and a = -9.8 ms^-2.

As v = u + at 

or, 0 = 49 - 9.8 x t

or, t = 49/9.8 = 5s

Total time = 5 + 5 = 10s

When lift is moving upwards with uniform velocity, initial velocity of ball will remain 49 ms^-1 only w.r.t., lift.Thus, the time taken by ball will be 10 s.