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 hoes the ball take to return to his hands ?

Explanation

When the lift is stationary :
Consider the motion of ball going upwards and then coming back to the hands of boy, i.e. displacement is zero.
`therefore S=0,u=49ms^(-1),g=-9.8ms^(-2)`
`S=ut+(1)/(2)gt^(2)`
`0=49+(1)/(2)(-9.8)t^(2)`
`rArr t=10s`
When lift starts moving, there is no change in relative velocity of ball w.r.t. boy and hence, time taken by the ball to return to boy.s hand will be 10 sec.