After climbing a wall of `3 m` height, a person of weight `W` jumps to the ground. If the body comes to a complete stop in `0.15 s`. After the feet touch the ground, calculate the average impulsive force in the vertical direction exerted by ground on the feet.

Options

A. `5W`
B. `5.21W`
C. `3W`
D. `6W`

Explanation

To solve the problem step by step, we need to calculate the average impulsive force exerted by the ground on the feet of a person who jumps from a height of 3 meters and comes to a stop in 0.15 seconds.

### Step 1: Calculate the speed just before hitting the ground
To find the speed of the person just before touching the ground, we can use the principle of energy conservation or kinematic equations. The potential energy at the height will convert into kinetic energy just before impact.

Using the equation:
\[ v = \sqrt{2gh} \]
where:
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height (3 m).

Substituting the values:
\[ v = \sqrt{2 \cdot 9.81 \cdot 3} = \sqrt{58.86} \approx 7.67 \, \text{m/s} \]

### Step 2: Calculate the change in momentum
The change in momentum (\(\Delta P\)) can be calculated using the formula:
\[ \Delta P = P_{\text{final}} - P_{\text{initial}} = 0 - mv \]
Since the person comes to a complete stop, the final momentum is 0. The initial momentum is \( mv \), where \( m \) is the mass of the person. Thus:
\[ \Delta P = 0 - mv = -mv \implies \Delta P = mv \]

### Step 3: Calculate the average impulsive force
The average impulsive force (\( F_{\text{average}} \)) can be calculated using the formula:
\[ F_{\text{average}} = \frac{\Delta P}{\Delta t} \]
where \(\Delta t\) is the time taken to stop (0.15 s). Substituting the values:
\[ F_{\text{average}} = \frac{mv}{0.15} \]

### Step 4: Substitute \( m \) with weight \( W \)
Since weight \( W \) is given by \( W = mg \), we can express mass \( m \) as \( m = \frac{W}{g} \). Substituting this into the force equation gives:
\[ F_{\text{average}} = \frac{(W/g)v}{0.15} \]

### Step 5: Substitute \( v \)
Now substitute \( v \) from Step 1:
\[ F_{\text{average}} = \frac{(W/g) \cdot 7.67}{0.15} \]

### Step 6: Calculate the average impulsive force
Substituting \( g \approx 9.81 \, \text{m/s}^2 \):
\[ F_{\text{average}} = \frac{(W/9.81) \cdot 7.67}{0.15} \]
Calculating this gives:
\[ F_{\text{average}} \approx 5.21 W \]

Thus, the average impulsive force exerted by the ground on the feet is approximately \( 5.21 W \).

### Final Answer
The average impulsive force in the vertical direction exerted by the ground on the feet is \( 5.21 W \).

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