A body is thrown vertically upwards from `A`. The top of a tower . It reaches the ground in time `t_(1)`. It it is thrown vertically downwards from `A` with the same speed it reaches the ground in time `t_(2)`, If it is allowed to fall freely from `A`. then the time it takes to reach the ground.
.

Options

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

Answer

C

Explanation

Suppose the body be projected vertically upwards from A with a speed `u_(0)`
Using equation `s=ut+((1)/(2))at^(92),` we get
For first case: `-h=u_(0)t_(1)-((1)/(2))g t_(1)^(2)` ..(i)
For second case : `-h=-u_(0)t_(2)-((1)/(2))g t_(2)^(2)` ..(ii)
`(i)-(ii)implies0=u_(0)(t_(2)+t_(1))t_(2)-((1)/(2))g t_(2)^(2)`
`impliesh=(1)/(2)g t_(1)t_(2)` ..(iv)
For third case: `u=0,t=?`
`-h=0xxt-((1)/(2))g t^(2) or h=((1)/(2))g t^(2)` ..(v)
Combining Eq. (iv) and (v) we get
`(1)/(2)g t^(2)=(1)/(2)g t_(1)t_(2) or t=sqrt(t_(1)t_(2))`