If you measure upwards from the ground then V is positive, the acceleration a is negative and the displacement s is given by
s = V t +^{1}/_{2} a t^{2}
where t is time measures from the instant that the body is released. The body returns to the ground when s = 0, so
V t +^{1}/_{2} a t^{2} = 0
Solving for t gives t = o and t = ^{2V}/_{a} so the body is at its maximum height at t = ^{V}/_{a} and this maximum height is
smax = s( ^{V}/_{a} ) = ^{ V2}/_{2a}
The question is then, what initial velocity will give a velocity of V at height smax, for that velocity will allow the body to rise another smax before starting to fall. Let U be this initial velocity. With this initial velocity, the velocity of the body at time t is given by U + at and the displacement by s = U t +^{1}/_{2} a t^{2}. Let t_{0} be the time when the body reaches a height of smax and then you want
V = U + a t_{0} and ^{ V2}/_{2a} = U t_{0} +^{1}/_{2} a t^{2}_{0}
I then used the first equation to write t_{0} = ^{(V  U)}/_{a} substituted this value into the second equation and solved for U. This gave me U = √2 V.
Penny
