physics question

A kinematics equation relates final speed, v, initial speed u, distance s and acceleration, a, by  v² = u² + 2*a*s  when a is constant.

In the situation here, v = 0 (it slows to zero at its highest point before coming back down), and a = -g (the acceleration of gravity, g, is constant and downward). So

0 = u² - 2*g*s

Add 2*g*s to both sides

2*g*s = u²

Divide both sides by 2*g

s = u²/(2*g)   ...(1)

To reach a height of 3s we need initial velocity of U, say, so we would have:

3s = U²/(2*g)

Replacing s using (1) we get

3*u²/(2*g) =U²/(2*g)

Multiply both sides by 2g

3u² = U²

Take the square root of both sides

U = u*√3

So the new initial velocity would have to be √3 times as big as the original.

Very nice, Alan......this explanation makes total sense........

Thank you alan, now its very clear