A man whose mass is `m` kg jumps vertically into air from the sitting position in which his centre of mass is at a height `h_(1)` from the ground. When his feet are just about to leave the ground his centre of mass is at height `h_(2)` from the ground and finally centre of mass rises to `h_(3)` above the ground when he is at the top of the jump. what is the average upward force exerted by the ground on him?

Options

A. (a)`(mg(h_(3)-h_(1)))/((h_(3)-h_(2)))`
B. (b)`(mg(h_(3)-h_(1)))/(h_(3))`
C. (c)`(mg(h_(3)-h_(1)))/((h_(2)-h_(1)))`
D. (d)none of these

Explanation

To find the average upward force exerted by the ground on a man who jumps vertically, we can follow these steps:

### Step 1: Understand the Problem
We have a man with mass `m` kg who jumps from a sitting position. His center of mass is at different heights during the jump:
- Initial height when sitting: \( h_1 \)
- Height just before leaving the ground: \( h_2 \)
- Maximum height reached: \( h_3 \)

### Step 2: Analyze the Motion
When the man jumps, he accelerates upwards until he reaches the maximum height \( h_3 \). At this point, his velocity becomes zero. We can use the kinematic equation to relate the heights and velocities:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity (0 at height \( h_3 \))
- \( u \) = initial velocity at height \( h_2 \)
- \( a \) = acceleration (which is \(-g\) due to gravity)
- \( s \) = displacement (\( h_3 - h_2 \))

### Step 3: Apply the Kinematic Equation
Since the final velocity \( v = 0 \) at height \( h_3 \), we can write:
\[ 0 = v^2 - 2g(h_3 - h_2) \]
Rearranging gives:
\[ v^2 = 2g(h_3 - h_2) \]

### Step 4: Calculate Change in Kinetic Energy
The change in kinetic energy as the man jumps from height \( h_1 \) to height \( h_2 \) is equal to the work done by the average force exerted by the ground. The work done can be expressed as:
\[ \text{Work done} = F \cdot (h_2 - h_1) \]
And the change in kinetic energy is:
\[ \Delta KE = \frac{1}{2} mv^2 \]

### Step 5: Set Up the Equation
Equating the work done to the change in kinetic energy:
\[ F(h_2 - h_1) = \frac{1}{2} mv^2 \]
Substituting \( v^2 \) from the previous step:
\[ F(h_2 - h_1) = \frac{1}{2} m(2g(h_3 - h_2)) \]
This simplifies to:
\[ F(h_2 - h_1) = mg(h_3 - h_2) \]

### Step 6: Solve for Average Force \( F \)
Rearranging gives:
\[ F = \frac{mg(h_3 - h_2)}{h_2 - h_1} \]

### Step 7: Include Weight of the Man
The total force exerted by the ground must also counteract the weight of the man:
\[ F - mg = \frac{mg(h_3 - h_2)}{h_2 - h_1} \]
Thus,
\[ F = mg + \frac{mg(h_3 - h_2)}{h_2 - h_1} \]
Factoring out \( mg \):
\[ F = mg\left(1 + \frac{h_3 - h_2}{h_2 - h_1}\right) \]

### Final Result
The average upward force exerted by the ground on the man is:
\[ F = mg \cdot \frac{h_3 - h_1}{h_2 - h_1} \]