Any object will accellerate due to gravity -9.81 m/s^2. By integrating this acceleration over time, one obtains the velocity at any time t.
v = gt + v0. In this case, the ball started at rest, so the initial velocity v0 is zero.
By integrating velocity over time, one obtains the position y at any time t.
y = 1/2gt^2 + y0. We know that y0, the initial height of the ball, is 50 m.
So, when does the ball hit the ground (y = 0). Solve for t:
0 = 1/2 (-9.81 m/s^2) t^2 + 50 --> 100s^2 = 9.81 t^2 --> t = 3.2 s
You can answer the second question by examining the energy of the system. The kinetic energy when the ball hits the ground is determined by it's velocity. 1/2 mv^2. The velocity at t = 3.2s is v = g*3.2 = 10.2 m/s.
Uk = 52*m J/kg
The problem states that the ball looses half it's energy to the impact. So the new kinetic energy is 26*m J/kg. How high will it go? All of the kinetic energy will go to potential energy according to the equation Up = mgh. So,
Uk = Up --> 52m = mgh --> h = 52/g = 5.3 meters.