A skydiver jumps from a helicopter at rest at an altitude of 3300 m above the gr

Question

A skydiver jumps from a helicopter at rest at an altitude of 3300 m above the ground. After some time of freefall, when her speed is 56 m/s and her altitude above the ground is 500 m, she opens the parachute. She lands with a speed of 6.0 m/s. Assuming the skydiver's mass was 60 kg., estimate: the net work done on the skydiver by the air and the parachute harness after the parachute opened the average net force exerted on her (by the air and the parachute harness) to slow her falling speed; and the work done on her by air resistance before the parachute opened. Model her as a particle.

Explanation / Answer

so at that time, her total energy(assuming height=0 i.e ground to be zero potential energy point)
=0.5*60*56^2+60*9.8*500=388080 J

when she reaches ground, kinetic energy=0.5*60*6^2=1800 J

so work done by air and parachute=388080-1800=386280 J

b)her net deceleration be a.
then 6^2-56^2=-2*a*500

a=3.1 m/s^2

hence net force on her=60*a=186 N

now as we can see, if force by air is F, then F-weight=186

F=774 N

c)when she started, her total energy=60*9.8*3300=1940400 J

when she reaches 500 m height and 56 m/s speed,

total energy=0.5*60*56^2+60*9.8*500= 388080 J

so work done by air resistance=1940400-388080=1552320 J

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