A stone dropped from a building of height `h` and it reaches after `t` second on the earth. From the same building if two stones are thrown (one upwards and other downwards) with the same speed and they reach the earth surface after `t_(1)` and `t_(2)` seconds, respectively, then

Options

A. `t=t_(1)-t_(2)`
B. `t=(t_(1)+t_(2))/(2)`
C. `t=sqrt(t_(1)t_(2))`
D. `t=sqrt(t_(1)^(2)-t_(1)^(2))`

Answer

c

Explanation

For stone dropped from rest we can write
`h=(1)/(2)gt^(2)`
For stone projected upward we can write
`h=-ut_(1)+(1)/(2)gt_(1)^(2)`
For stone projected downward we can write
`h=ut_(2)+(1)/(2)gt_(1)^(2)`
multiplying equation ii with `t_(2)` and equation iii with `t_(1)` and then adding both we get the following:
`h(t_(1)-+t_(2))=(1)/(2)gt_(1)t_(2)(t_(1)+t_(2))`
`rArr h=(1)/(2)gt_(1)t_(2)`
Comparing equation (i) and (iv) we get
`t =sqrt(t_(1)t_(2))`