A ball is thrown from the top of a tower in vertically upward direction. Velocity at a point h m below the point of projection is twice of the velocity at a point h m above the point of projection. Find the maximum height reached by the ball above the top of tower.

Options

A. ` 2h`
B. ` 3h`
C. ` (5//3) h`
D. ` (4//3)h`

Answer

C

Explanation

`H=(u^(2))/(2g)`, given `v_(2)=2v_(1)` (i)
`A` to `B`: `v_(1)^(2) =u^(2)-2 g t` (ii)
`A` to `C`: `v_(2)^(2)=u^(2)-2g (-h)` (iii)
Solving (i) ,(ii) and (iii), we get the value `u^(2)` as `10gh//3 ` and then we get the value of `H` by using
.
`H= (u^(2))/(2g)`.