The acceleration of a ball is constant; it is called gravity acceleration and is denoted as `g approx 10 m/s^2.` It is directed downwards.
The upward vertical speed of a ball uniformly decreases until it becomes zero, and a ball reaches its maximum height at that moment. Then the vertical speed becomes downward and increases uniformly. The general formula is `V(t) = V_0 - g t,` where `V_0` is the initial upward speed. So the time needed to reach a maximum height is `t_1 = V_0/g.`
The formula for height is `H(t) = V_0 t - (g t^2)/2.` Thus the maximum height is `H_(max) = H(t_1) = V_0^2/(2g).` It is given, so we can find `V_0 = sqrt(2gH_(max)).`
A ball falls at the moment `t_2gt0` at which `H(t_2) = 0,` so `t_2 = (2V_0)/g.` Substitute the formula for `V_0` and obtain `t_2 = 2sqrt((2H_(max))/g) approximately 2.8(s).` This is the answer.