# Stone Throw Kinematics Question

1. Jan 30, 2010

### Procrastinate

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

v2 = u2 + 2as

3. 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.

2. Jan 30, 2010

### 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.