Verified answer

Assuming a perfectly elastic collision with the floor (i.e. no energy is lost due to deforming the ball, the floor, etc.), the ball must return to the height where its speed is zero. At that height above teh floor, the ball's potential energy will be equal to the ball's kinectic energy just before it struck the floor.

So KE = PE --> 1/2mv^2 = mgh

or solving for h;

h = 1/2 v^2/g =0.5*(9.97 m/s)^2/(9.8m/s^2) = 5.07 m