Peter, a 100 kg basketball player, lands on his feet after completing a slam dunk and then immediately jump up again to celebrat
e his basket. When his feet first touch the floor after the dunk, his velocity is 5 m/s downward; when his feet leave the floor 0.50 s later, as he jumps back up, his velocity is 4 m/s upward. What is the average reaction force exerted upward by the floor on Peter during this 0.50 s?
Using the equation; TE = 1/2mv^2(1+2); where k = 2/5 for a solid sphere; V is the velocity, and m is the mass. Total energy = 0.5 × 21 × 8² (7/5) = 940.8 J The rotational kinetic energy of the sphere is 940.8 J