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.
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.
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
To find the height from which the student fell, we need to use the equations of motion and the force exerted on the student.
Given:
Mass of the student (m) = 63 kg
Time taken to come to rest (t) = 0.0120 s
Force exerted by the ground (F) = +17000 N
We can calculate the acceleration (a) experienced by the student using Newton's second law of motion:
F = m * a
Rearranging the equation, we have:
a = F / m
Substituting the values:
a = 17000 N / 63 kg
Find the acceleration (a).
a = 270.654 m/s^2
Next, we can use the equation of motion to find the height (h) from which the student fell, assuming the student was initially at rest:
h = (1/2) * a * t^2
Substituting the values:
h = (1/2) * 270.654 m/s^2 * (0.0120 s)^2
Find the height (h).
h = 0.19526 m
Therefore, the student fell from a height of approximately 0.19526 meters.
To find the height from which the student fell, we can use the equation for the average force during a collision:
Average Force = (Change in momentum) / (Time of collision)
First, let's find the change in momentum (Δp).
Since the student falls freely from rest, his initial momentum is zero.
The final momentum is given by:
Pf = m * vf
where m is the mass of the student and vf is the final velocity.
Since the student comes to rest, his final velocity (vf) is zero.
Therefore, Δp = Pf - Pi = m * vf - m * vi = m * vf
Now, we can use the equation for average force:
Average Force = Δp / (Time of collision)
Substituting the given values:
17000 N = (m * vf) / (0.0120 s)
Now, solve for vf:
vf = (Average Force * Time of collision) / m
vf = (17000 N * 0.0120 s) / 63 kg
vf ≈ 324.76 m/s
To find the height from which the student fell, we can use the equation:
vf^2 = vi^2 + 2g * h
where vi is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Since the student falls freely, vi is also zero.
Therefore, vf^2 = 2g * h
Substituting the known values:
(324.76 m/s)^2 = 2 * 9.8 m/s^2 * h
Now, solve for h:
h = vf^2 / (2g)
h = (324.76 m/s)^2 / (2 * 9.8 m/s^2)
h ≈ 546.06 m
Therefore, the student fell from a height of approximately 546.06 meters.