start: PE=mgh = 73.9 * 9.81 * 2570
KE = 1/2 mv^2 = 1/2 * 73.9 * (112 km/hr * 1000m/km * 1hr/3600s)^2
end: PE = mgh = 0
KE = 1/2 mv^2 = 1/2 * 73.9 * 5.21^2
A parachutist with mass 73.9 kg jumps from an airplane traveling at a speed 𝑣𝑖=112 km/hr at a height 𝐻=2570 m. He lands with a speed of 𝑣𝑓=5.21 m/s. What is the change in mechanical energy of the earth-parachutist system from just after the jump to just before landing?
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1 answer