A student (m = 68 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.
(Average force during impact)*(time)
= (momentum at impact)
Compute the momentum at impact and use the associated velocity V to get the initial height, H.
V = sqrt(2gH)
H = V^2/(2g)
We can solve this problem using the equation for average force:
\[ F_{avg} = \frac{{\Delta p}}{{\Delta t}} \]
Where:
- \( F_{avg} \) is the average force exerted on an object,
- \( \Delta p \) is the change in momentum of the object,
- \( \Delta t \) is the time interval over which the force is applied.
Since the student falls freely, his initial momentum is zero. Therefore, the change in momentum is equal to the final momentum:
\[ \Delta p = m \cdot v_f \]
Where:
- \( m \) is the mass of the student (68 kg in this case),
- \( v_f \) is the final velocity of the student.
From the problem statement, we know that the student comes to rest in a time of 0.04 s. The final velocity can be calculated using the following equation of motion:
\[ v_f = v_i + a \cdot t \]
Where:
- \( v_i \) is the initial velocity of the student (which is zero since he starts from rest),
- \( a \) is the acceleration, and
- \( t \) is the time of the collision (0.04 s in this case).
We also know that acceleration is given by:
\[ a = \frac{{F_{avg}}}{{m}} \]
So, we can substitute the values into the equations to solve for the final velocity:
\[ v_f = 0 + \left( \frac{{F_{avg}}}{{m}} \right) \cdot t \]
Finally, we can solve for the height the student fell from using the equation for gravitational potential energy:
\[ PE = m \cdot g \cdot h \]
Where:
- \( PE \) is the potential energy,
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s\(^2\)),
- \( h \) is the height.
Since the student started from rest, his initial kinetic energy is zero. Therefore, the potential energy gained during the fall is equal to the change in kinetic energy during the collision:
\[ PE = \Delta KE \]
\[ m \cdot g \cdot h = \frac{1}{2} \cdot m \cdot v_f^2 \]
Now we can solve for the height:
\[ h = \frac{{v_f^2}}{{2 \cdot g}} \]
We have all the necessary equations to solve for the height the student fell from. Let's substitute the given values into the equations to find the answer.
To find the height from which the student fell, we need to use the equations of motion for free fall and the relationship between force, mass, and acceleration.
First, let's list the known values:
Mass of the student (m) = 68 kg
Time taken to stop (t) = 0.04 s
Force exerted by the ground (F) = 18000 N
We'll use the equation for acceleration:
a = F / m
Rearranging the equation, we get:
a = (final velocity - initial velocity) / t
Since the student falls freely without any initial velocity, the initial velocity is 0 m/s. The final velocity is also 0 m/s since the student comes to rest. So, the equation becomes:
a = (0 - 0) / t
a = 0 / t
a = 0 m/s^2
Now, we can use the equation of motion to find the height:
s = ut + (1/2)at^2
Since the initial velocity (u) is 0 m/s and the acceleration (a) is 0 m/s^2, the equation simplifies to:
s = 0t + (1/2)(0)(t^2)
s = 0
Therefore, from the equations of motion, we can see that the height from which the student fell is 0 meters.