Hello!

The speed of a falling ball is `g*t` downwards, where g is the gravity acceleration. The height of a falling ball is

`H(t)=H_0-g(t^2)/(2),`

where `H_0` is the initial height.

A ball strikes the ground when H(t)=0, so

`t_1=sqrt((2H_0)/g),`

and the speed will be `V_1=t_1g=sqrt(2H_0g) approx 17(m/s).` This is the answer to the first part.

The kinetic energy before the strike is `mV_1^2/2,` at the end of the re-bouncing it will be zero. The change of the potential energy will be `mgH_1` , where `H_1` is the maximum height after the re-bouncing. So

`mgH_1=0.4m(V_1)^2/2,` or

`H_1=0.4(V_1)^2/(2g)=0.4H_0=6(m).`

This is the answer for the second part.

That said, the first part could be solved quickier using energy considerations (potential energy becomes kinetic), and the second - considering the potential energy before the fall :)