Question

A very large pumpkin with a mass of 31.0 kg is released from rest from a certain height and hits the ground. During the collision with the ground, the pumpkin comes to rest in a time of 0.040s. The average force exerted on the pumpkin by the ground is +12,000 N, where the upward direction is taken to be the positive direction.
a. With what velocity did the pumpkin hit the ground? Assume that the only force acting on the pumpkin during the collision is that due to the ground.
b. From what height did the pumpkin fall?

Answer 1

(a) Impulse = Favg x time of impact = change in momentum

12000 x .040 = change in momentum = m (Vf - Vi) (here Vf = 0)

480 = 31 x Vi

Vi = 15.48 m/s

(b) Using energy conservation

Initial KE + Initial PE = FInal KE + Final PE

0 + mgh = (KE)f + 0

Initial KE is '0' beacuase it released from rest
FInal PE is '0' beacause it is on groung 'h = 0'

So
mgh = 1/2 mv^2

v = 15.48 m/s

Putting the values

31 x 9.8 x h = .5 x 31 x (15.48)^2

303.8 x h = 3714.2712   

h = 12.226 meter