A ball is thrown vertically upwards with a speed of `10 m//s` from the top of a tower 200 m height and another is thrown vertically downwards with the same speed simultaneously. The time difference between them on reaching the ground is `(g=10 m//s^(2))`

Options

A. 12
B. 6
C. 2
D. 1

Answer

C

Explanation

`s = ut + (1)/(2) a t^(2)`

For first particle
`-200 = 10 xx t_(1) - (1)/(2) xx 10 xx t_(1)^(2)`
`-40 = 2t_(1) - t_(1)^(2)`
`t_(1)^(2) - 2 t_(1) - 40 = 0`
`t_(1) = (2 +_ sqrt(164))/(2)` (`+v` sign is considered)
`:. t_(1) = (2 + sqrt(164))/(2)`
For second particle
`200 = 10 t_(2) + (1)/(2) 10 t_(2)^(2)`
`:. t_(2) = (-2 + sqrt(164))/(2)` (considering only `+ ve` sgn)
`:. Delta t = t_(1) - t_(2) = 2s`