Question

A boy standing on a stationary lift ( open from above ) thrown 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 hoes the ball take to return to his hands ?

Answer 1

Taking vertical upwward direction as the positive direction of x-axis.
When lift is stationary , consider the motion of the ball going vertically upwards and coming down to the hands of the body, we have ` u=49 ms^(-1)`, a =- 9.8 ^(-1), t= ?, x-x_0 = S= 0`
As, S=ut + 1/2 at^2`
:. ` 0= 49 t + 1/2 (- 9.8 ) t^2 or 49 t = 4.8 T2 or t= 49 //4.9 = 10 seconds`
When lift stars moving with uniform speed
As the lift starts moving upwards with uniform speed of ` 5 ms^(-1)`, there is no change in the relative velocity or the ball w.r.t. the boy which remains ` 49 ms^(-1)`, Hence, even in this case, the ball will return to the boy `s hand after ` 10 second`.