Question

A 82 kg person has his center of gravity 1.0 meters above the ground in a standing position (height that person comes off the ground during a jump). During a squat jump, he crouches down so that his centre of gravity is 0.3 meters above the ground. As he jumps, he exerts an average force of 1258 Newtons against the ground. You can think of this average exerted force as the average normal force or average ground reaction force being read by the force plate during the takeoff phase of the jump. Determine how high the person will jump in meters using the concepts of Work and Energy.

Answer 1

here we use Work energy theoam Wgravity + Whormed force = (kEf-KE;) It height jumped by person is h gravity = -migh W normed force = Fd , KE;=u, kEf=0 e da 1-0.3= 0.7m. - migh +Fd=0 - 82x9.81 xh = - 1258 X 0.7 h = 1.0947 m initial height of centre of gravity = 0.3m le Height from ground of centre of gravity H = 0.3th TH= 1.3947 m Only height of sump = 0.3947m