A student (m = 66 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.04 s. The average force exerted on him by the ground is +18000 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.
Tried finding an equation that works for this but can't figure it out. I know it has to do with impulse-momentum theorem but I can't figure out how to apply the theory.
5.3
To solve this problem, we can use the concept of impulse-momentum theorem to find the velocity of the student just before hitting the ground. Then, we can use this velocity and the time of impact to calculate the height from which the student fell.
Let's break down the steps to solve the problem:
1. Using the impulse-momentum theorem:
The impulse-momentum theorem states that the change in momentum of an object is equal to the net impulse acting on it. Mathematically, it can be expressed as:
J = Δp
where J is the impulse and Δp is the change in momentum.
In this case, we can assume that the only force acting on the student during the collision is that due to the ground. Therefore, the net impulse acting on the student is equal to the average force exerted by the ground multiplied by the time of impact:
J = F * Δt
Substituting the values given in the problem, we have:
J = 18000 N * 0.04 s
2. Finding the change in momentum:
The change in momentum is equal to the final momentum minus the initial momentum:
Δp = p_f - p_i
Since the student comes to rest, the final momentum (p_f) is 0 kg·m/s. Therefore, the change in momentum is simply equal to the initial momentum (p_i):
Δp = p_i
3. Calculating the velocity just before hitting the ground:
The momentum of an object is given by:
p = m * v
where p is the momentum, m is the mass, and v is the velocity.
Therefore, we can write the initial momentum as:
p_i = m * v_i
Now, substituting the values in the equation for the impulse-momentum theorem:
J = p_i
J = m * v_i
4. Solving for v_i:
Rearranging the equation, we have:
v_i = J / m
Substituting the values from step 1:
v_i = (18000 N * 0.04 s) / 66 kg
Simplifying the equation:
v_i = 10.91 m/s
5. Calculating the height:
When an object falls freely, its final velocity can be calculated using the equation:
v_f^2 = v_i^2 + 2gh
where v_f is the final velocity (0 m/s in this case), v_i is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Substituting the values into the equation and rearranging gives:
0 = (10.91 m/s)^2 + 2 * 9.8 m/s^2 * h
Simplifying the equation:
h = - (10.91 m/s)^2 / (2 * 9.8 m/s^2)
h = -5.95 m
The negative sign indicates that the height is below the ground level. Therefore, disregarding the sign, the student fell from a height of approximately 5.95 meters.
Remember to always double-check the units and perform the calculations accurately.