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 +16000. 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.
2006-11-06 16:18:46 · 3 answers · asked by Alan l 1 in Science & Mathematics ➔ Physics
This is an impulse/momentum problem. Impulse is ∫f*dt[0 - t1] it must equal the change in momentum. Impulse is given by Favg*t, in your case 0.012*16000 N*sec Therefore the momentum of the student on impact was 0.012*16000 = m*v. m is the mass of the student, 63kg. Therefore his velocity on impact was v =0.012*16000/63. He fell under acceleration g = 9.8m/sec^2. Velocity after falling time t is g*t. Distance traveled under uniform acceleration is h=.5*g*t^. Use the velocity to get t, then compute h.
2006-11-06 16:35:48 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋
V at ground zero = Ft/m
Height H = (V^2)/2g
Plug in your numbers and enjoy the answer......
2006-11-07 00:24:22 · answer #2 · answered by Steve 7 · 0⤊ 0⤋
errr..the only thing that i know is that it falls due to gravity which is 10 N.the rest,omg!i forgot.this is form 4 syllabus right?
2006-11-07 00:26:10 · answer #3 · answered by leeyawna 2 · 0⤊ 0⤋