(1/2) m v^2 = m g h
so
v = sqrt( 2 g h ) = sqrt (2 * 9.81 * 1000 ) = 140 m/s
Potential energy at top = m g h = 70 * 9.81 *1000 = 686,700 Joules
Kinetic energy at ground = (1/2) m v^2 = 35 * 62^2 = 134,540 Joules
lost to friction= 552,160 Joules warming air
By landing properly and on soft ground (and by being lucky!), humans have survived falls from airplanes when, for example, a parachute failed to open, with astonishingly little injury. Without a parachute, a typical human eventually reaches a terminal velocity of about 62.0 m/s. Suppose the fall is from an airplane 1000 m high.
How fast would a person be falling when he reached the ground if there were no air drag?
(in m/s)
If a 70.0 kg person reaches the ground traveling at the terminal velocity of 62.0 m/s, how much mechanical energy was lost during the fall?
1 answer