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 goes the ball take to return to his hands ?

Explanation

Case:1
When lift is stationary,
The ball thrown by boy with maximum velocity `49 ms^(-1)` comes back tgo boy after reaching initial velocity.
`therefore` Initial velocity `u = 49 ms ^(-1)`
`a = g= - 9.8 ms ^(-2)`
`t = ? S = 0m`
From third equation of motion,
`s = ut + 1/2 at ^(2)`
`0=49 t + 1/2 xx (-9.8) t ^(2)`
`therefore 4.9 t ^(2) = 49t`
`therefore t = (49)/(4.9) `
`therefore t = 10 s`
Second method :
Suppose, time taken to reach maximum height is t,
From first equation of motion,
`v = u + at`
`0 = 49 + (-9.8)t`
`therefore 9.8 t = 49`
`therefore t = (49)/(9.8) =5s`
Time taken to reach maximum height = time taken to return back
`therefore 5 = t`
`therefore ` Total time taken `T = 2t = 2 xx 5 = 10 s`
Case : 2
WHen lift moves upwards with constant speed. In this case when the lift moves upwards with the coonstant speed of `5 ms ^(-1),` then the relative velocity of ball will be 49 m/s. Hence, ball will return back in 10 s.