A parachutist of weight ‘ w ’ strikes the ground with his legs fixed and comes to rest with an upward acceleration of magnitude 3 g . Force exerted on him by ground during landing is

Options

A. w
B. 2 w
C. 3 w
D. 4 w

Answer

To solve the problem, we need to analyze the forces acting on the parachutist when he strikes the ground and comes to rest with an upward acceleration of magnitude \(3g\).

### Step-by-Step Solution:

1. **Identify the Weight of the Parachutist:**
The weight of the parachutist is given as \(w\). This weight can be expressed in terms of mass \(m\) and gravitational acceleration \(g\):
\[ w = mg \]

2. **Determine the Downward Acceleration:**
The parachutist comes to rest with a downward acceleration of \(3g\). This means that while he is decelerating, he experiences an effective downward force due to this acceleration.

3. **Calculate the Force Due to Downward Acceleration:**
The force exerted by the parachutist due to this downward acceleration can be calculated using Newton's second law:
\[ F_{\text{down}} = m \cdot a = m \cdot 3g = 3mg \]

4. **Calculate the Weight Force:**
The weight force acting on the parachutist is:
\[ F_{\text{weight}} = mg = w \]

5. **Determine the Total Force Exerted on the Ground:**
When the parachutist lands, the total force exerted on the ground (normal force \(N\)) is the sum of the weight force and the force due to the downward acceleration:
\[ N = F_{\text{weight}} + F_{\text{down}} = mg + 3mg = 4mg \]

6. **Substitute the Weight into the Equation:**
Since \(mg = w\), we can substitute this into our equation for the normal force:
\[ N = 4mg = 4w \]

7. **Final Answer:**
Therefore, the force exerted on him by the ground during landing is:
\[ \boxed{4w} \]