The speed of a falling ball is `g*t` downwards, where g is the gravity acceleration. The height of a falling ball is
where `H_0` is the initial height.
A ball strikes the ground when H(t)=0, so
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
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 :)