Question

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

Answer 1

Correct Answer - C
Maximum height reached, `H = (u^2)/(2 g)`
Given `v_2 = 2v_1 rArr (v_1)/(v_2) = (1)/(2)`…(i)
Motion from `A and B` :
`v_1^2 = u^2 - 2gh` …(ii)
Motion from `A` to `C` :
`v_2^2 = u^2 - 2gh (-h)`…(iii)
From (ii)/(iii), `((v_1^2)/(v_2))^(2) = (u^2 - 2gh)/(u^2 + 2gh) = (((u^2)/(2g) - h))/(((u^2)/(2g) + h))`
`(1)/(4) =((H - h))/((H + h)) rArr 4 (H - h) = H + h`
`(H = (u^2)/(2 g))`
`rArr 3H = 5h rArr H = (5)/(3) h`.