- #1
kirby27
- 32
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a 64 kg parachutist falling vertically at a speed of 6.3 m/s impacts the ground, which brings him to a complete stop in a distance of .92 m. assuming constant acceleration after his feet first touch, what is the average force exerted on the parachutist by the ground?
i know:
Vf=0
Vi=6.3
delta Y = .92
i used (Vfy)^2=(Viy)^2+2a(deltaY)
and found a = -21.57
i then used F=ma: F= (65)(-21.57)
and got -1402 N
is this right?
i know:
Vf=0
Vi=6.3
delta Y = .92
i used (Vfy)^2=(Viy)^2+2a(deltaY)
and found a = -21.57
i then used F=ma: F= (65)(-21.57)
and got -1402 N
is this right?