Answer:
If he throws the ball with speed V then (vertical speed)
V / g is the time to reach maximum height (V2 - V1) / a
The total time to again reach the ground is then
T = 2 V / g
H = 1/2 g t^2 where t = V / g
H = 1/2 g (V / g)^2 = V^2 / (2 g)
This is the familiar formula
V = (2 g H)^1/2 for the initial upwards speed
