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 bail 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 hands?

Answer 1

Initial velocity n = 49 ms^-1

Acceleration g = -9.8 ms²

Displacement s = 0

s = ut + 1/2 at²

0 = 49 t + 1/2 (-9.8) t²

⇒ t = 0 or t = 10 s

t = 0 represents initial time. The required ‘t’ here is t = 10 s. In the case where the lift also moves, the relative velocity of the ball with respect to the boy remains the same, i.e, 49 ms^-1. The relative displacement is also 0. Hence the time required in this case is also 10 s.