A ball is fallen freely from top of tower of height 100 m and another ball is thrown upwards with 50 m/s at the same time when they will cross each other ? `(g = 10 m//s^(2))`

Options

A. 1 s
B. 2 s
C. 3 s
D. 4 s

Answer

B

Explanation

For the ball falling freely `v _(0) = 0, d = 100 m`
`therefore` Distance covered in .t. time
`d _(1) = v _(0)t + 1/2 g t^(2)`
`d _(1) = 0 + 1/2 (10) t ^(2) = 5t ^(2) " "…(1)`
Initial velocity of the ball thrown upwards is `v _(0).=50m//s` and it meets the first ball at `(100 -d _(1)`)m distance
`therefore 100 -d _(1) = v _(0) .t = 1/2 g t ^(2)`
`therefore 100 - d _(1) = 50 t - 1/2 (10) t ^(2) `
`therefore 100 -d _(1) = 50 t -5t ^(2)`
From equation (1),
`therefore 100-5t ^(2) = 59 t - 5t ^(2)`
`therefore 50 t = 100`
`therefore t = (100)/(50)`
`therefore t =2s`