Stone Throw Kinematics Question

[Procrastinate]

Homework Statement

A stone is projected vertically upwards from the top of a tower. It is found that its speed at a distance of x meters below the point of projection is twice its speed at the same distance above that point. Show that the stone rises a distance 5x/3metres above the top of the tower.

Homework Equations

v² = u² + 2as

The Attempt at a Solution

Let the velocity from the projection point upwards be v₁
Let the velocity from the projection point downwards be v₂

v₁ = u² - 19.6x
v₂ = u² + 19.6x

Since V₁ = 2v₂

u² - 19.6x = 2u² + 39.2x

-58.8x = u²

I am stumped.

 

[rl.bhat]

When the stone reaches the maximum height h, its velocity is equal to zero.
From that height if the stone is dropped, its velocity at a distance (h-x) is
v^2 = 2*g*(h-x). ...(1)
According to the problem, its velocity at a distance (h+x) is 2v and it is equal to
(2v)^2 = 2*g*(h+x)...(2)
Divide eq.2 by eq.1 and solve for h in terms of x.