A student (m = 64 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.016 s. The average force exerted on him by the ground is +19000 N. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.
2007-06-18 15:12:22 · 4 answers · asked by iceosiris117 1 in Science & Mathematics ➔ Physics
id say nearly 540 feet.
2007-06-18 15:20:54 · answer #1 · answered by devilz_krypt 2 · 0⤊ 0⤋
By the momentum-impulse theorem, we know that the change in momentum equals the impulse:
dp = F*t
dp = (19000 N)*(0.016 s)
dp = 304 kg*m/s
Since he came to rest, his change in momentum is equal to his momentum just before hitting the ground. His velocity just before hitting was:
p = m*v
304 kg*m/s = (64 kg)*v
v = 4.75 m/s
His kinetic energy was therefore
T = 1/2*m*v^2
T = 1/2*(64 kg)*(4.75 m/s)^2
T = 722 joules
Since he had 722 joules of kinetic energy just before hitting the ground, he must have started his fall with 722 joules of potential energy. The potential energy in a uniform gravity field is:
U = m*g*h
722 = (64 kg)*(9.81 m/s^2)*h
h = 1.15 m
He fell from a height of 1.15 meters.
2007-06-18 22:24:27 · answer #2 · answered by lithiumdeuteride 7 · 0⤊ 0⤋
the above answer is correct but he done it in a very lengthy way. he is right up to he finds the velocity v=4.75 m/s
now just use the relation ,
v^2=(2gh)
where v is velocity
g is acc. due to gravity
h is the height
so,
(4.75)^2=2X10Xh
=>h=((4.75)^2)/20
=>h=1.128~1.13 (ans)
2007-06-20 04:21:59 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Do your home work yourself
2007-06-18 22:19:45 · answer #4 · answered by Sarvo 2 · 1⤊ 0⤋