A ball thrown upward from the top of a tower with speed `v` reaches the ground in `t_(1)` sec. If this ball is thrown downward from the top of the same tower with speed `v` it reaches the ground in `t_(2)` sec. In what time, the ball shall reach the grouned, if it is allowed to fall freely under gravity from the top of the tower?

Options

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

Explanation

(c) `h=-vt_(1)+1/2"gt"_(1)^(2)` or `h/(t_(1))=-v+1/2"gt"_(1)`………(i)
`h=vt_(2)+1/2"gt"_(2)^(2)` or `(-h)/(t_(2))=-v+1/2"gt"_(2)`……..(ii)
`:. h/(t_(1))+h/(t_(2))=1/2 g(t_(1)+t_(2))`
or `h=1/2"gt"_(1)t_(2)`
From falling under gravity form the top of the tower
`:. 1/2"gt"_(1)t_(2)=1/2"gt"^(2)impliest=sqrt(t_(1)t_(2)`