A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to `49 m s^-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 m s^-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 ?

Explanation

When lift is stationary, consider the motion of the ball going vertically upwards and coming down to the hands of the boy, we have given `: u = 49 ms ^(-1) , a =- 9.8 ms ^(-2) s = 0`
Using relation ` : s = ut + (1)/(2) at ^(2) implies 0 = 49 t - (1)/(2) (9.8) t ^(2) or, t = ( 49)/(4.9) = 10 s.`
When lift is moving upwards with the uiform velocity, initial velocity, of ball will rmain `49 ms ^(-1)` w.r.t. lift. So the time taken by ball will be 10 s.