In order to make a jump straight up, a `60 kg` player starts the motion crouched down at rest. He pushes hard against the ground, raising his centre of mass by a height `h_(0) = 0.5 m`. Assume that his legs exert a constant force `F_(0)` during this motion. At this point, where his centre of mass has gone up by `h_(0)` his feet leave the ground and he has an upward velocity of `v`. Centre of mass of his body rises further by `h = 0.8 m` before falling down [Take `g = 10 m//s^(2)`]
(a) Find `v`.
(b) Find the normal force applied by the ground on his feet just before he left the ground.

Answer

(a) `4 m//s`
(b) `1560 N`