A 1 kg basketball is dropped from the fifth floor of an apartment building (15 meters high) onto the sidewalk?

Question

Below. The collision dissipates 50% of the energy as heat. How high does the ball bounce?Answer in meters.

Answer 1

The total energies (TE) at three way-points in the ball's trajectory are:

At release point: TE(H) = mgH
At bounce: TE(0) = KE(0) + WE(0) = KE(0) + 1/2 mgH
At bound height: TE(h) = mgh

Where PE(H) = mgH the potential energy at H = 15 m, 1/2 mv^2 = KE(0) the rebound kinetic energy, 1/2 mgH = 50% energy lost to heat, mgh = PE(h) potential energy when the ball reaches h height, the answer you are looking for.

From the conservation of energy TE(H) = mgH = KE(0) + 1/2 mgH = TE(0); so that KE(0) = 1/2 mgH.

Then TE(h) = mgh = 1/2 mgH = TE(0) from the conservation of energy where half of it was lost to heat.

Thus h = 1/2H is the bounce height ~ 7.5 meters

Answer 2

use 1/2 m v^2=mgh