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 0.010 s. The average force exerted on him by the ground is +18,000 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.

Options

A. 0.15 m
B. 0.21 m
C. 0.30 m
D. 0.42 m

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the change in momentum
The average force exerted on the student during the collision with the ground is given as \( F = 18,000 \, \text{N} \). The time duration of the collision is \( \Delta t = 0.010 \, \text{s} \).

Using the impulse-momentum theorem:
\[ F \Delta t = m \Delta v \]
where \( m \) is the mass of the student and \( \Delta v \) is the change in velocity. Rearranging gives:
\[ \Delta v = \frac{F \Delta t}{m} \]

### Step 2: Substitute the known values
Substituting the known values into the equation:
- Mass \( m = 63 \, \text{kg} \)
- Force \( F = 18,000 \, \text{N} \)
- Time \( \Delta t = 0.010 \, \text{s} \)

\[ \Delta v = \frac{18,000 \, \text{N} \times 0.010 \, \text{s}}{63 \, \text{kg}} = \frac{180}{63} \approx 2.857 \, \text{m/s} \]

### Step 3: Calculate the velocity just before impact
Since the student falls freely from rest, the velocity just before impact \( v \) is equal to \( \Delta v \):
\[ v \approx 2.857 \, \text{m/s} \]

### Step 4: Use the kinematic equation to find the height
Using the kinematic equation for free fall:
\[ v^2 = 2gh \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)) and \( h \) is the height. Rearranging gives:
\[ h = \frac{v^2}{2g} \]

### Step 5: Substitute the values to find height
Substituting the values:
\[ h = \frac{(2.857 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2} = \frac{8.163}{19.6} \approx 0.416 \, \text{m} \]

### Step 6: Final answer
Thus, the height from which the student fell is approximately:
\[ h \approx 0.42 \, \text{m} \]