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 hands ?

Explanation

Taking vertical upward direction as the positive direction of x-axis.
When lift is stationary, consider the motion of the ball going vertically upwardes and coming down to the hands of the body. We have
`u=49m//s, a =9.8m//s^(2), t =?" "x-x_(0)=S=0`
As `S=ut+1/2at^(2)`
`0=49t+1/2(-9.8)t^(2)" (Or) "49t=4.9t^(2)" (Or)"`
`t=49//4.9=10sec`
When lift starts moving with uniform speed. As the lift starts moving upwards with uniform. Speed of 5 m/s, there is no change in the relative velocity of the ball w.r.to the boy which remains 49 m/s. Hence even in the case, the ball will return to the boys hand after 10 sec.