To solve the problem of a person with a mass of 70 kg jumping from a height of 3.0 m, we will follow these steps:
### Step 1: Calculate the final velocity just before hitting the ground
Using the equation of motion:
\[
v^2 = u^2 + 2gh
\]
where:
- \( v \) = final velocity just before impact,
- \( u \) = initial velocity (0 m/s, since the person jumps from rest),
- \( g \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( h \) = height (3.0 m).
Substituting the values:
\[
v^2 = 0 + 2 \times 10 \times 3
\]
\[
v^2 = 60
\]
\[
v = \sqrt{60} \approx 7.75 \, \text{m/s}
\]
### Step 2: Calculate the impulse on the person by the ground
Impulse is given by the change in momentum:
\[
\text{Impulse} = \Delta p = m(v - 0)
\]
where:
- \( m = 70 \, \text{kg} \),
- \( v \approx 7.75 \, \text{m/s} \).
Calculating the impulse:
\[
\text{Impulse} = 70 \times 7.75 \approx 542.5 \, \text{Ns}
\]
### Step 3: Calculate the average force during the impact
The average force can be calculated using the work-energy principle:
\[
F_{\text{average}} \cdot d = \Delta KE
\]
where:
- \( d \) = distance over which the force acts (let's consider two cases: 1 cm and 50 cm),
- \( \Delta KE = \frac{1}{2} mv^2 \).
Calculating the change in kinetic energy:
\[
\Delta KE = \frac{1}{2} \times 70 \times (7.75)^2
\]
\[
\Delta KE = \frac{1}{2} \times 70 \times 60 \approx 2100 \, \text{J}
\]
#### Case 1: For \( d = 0.01 \, \text{m} \) (1 cm)
Using the formula:
\[
F_{\text{average}} = \frac{\Delta KE}{d}
\]
\[
F_{\text{average}} = \frac{2100}{0.01} = 210000 \, \text{N}
\]
#### Case 2: For \( d = 0.50 \, \text{m} \) (50 cm)
\[
F_{\text{average}} = \frac{2100}{0.50} = 4200 \, \text{N}
\]
### Summary of Results
1. Final velocity just before impact: \( \approx 7.75 \, \text{m/s} \)
2. Impulse on the person by the ground: \( \approx 542.5 \, \text{Ns} \)
3. Average force during impact:
- For 1 cm: \( 210000 \, \text{N} \)
- For 50 cm: \( 4200 \, \text{N} \)
