Ground-applied impulse = momentum at impact
17,000*0.0120 = 204 kg*m/s = M*V
V = 3.23 m/s
(1/2)V^2 = g H
(Energy is conserved while the student is falling)
Solve for H, the height of the drop.
A student (m = 63 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.0120 s. The average force exerted on him by the ground is +17000. N, where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.
2 answers
Ground applied impulse =momentum at impact .
17000× 0.0120=mv
204=(63)v
:.v=3.23
Then use the equation of motion
Y= vt +1/2gt^2
17000× 0.0120=mv
204=(63)v
:.v=3.23
Then use the equation of motion
Y= vt +1/2gt^2
2 answers