A body of mass m when released from rest from a height h, hits the ground with speed `sqrt(gh)` . Find work done by air resistance force.

Options

A. `-mgh`
B. `mgh `
C. `-(mgh)/(2)`
D. `(mgh)/(2)`

Explanation

To solve the problem of finding the work done by air resistance force on a body of mass \( m \) released from a height \( h \) and hitting the ground with speed \( \sqrt{gh} \), we can follow these steps:

### Step 1: Identify the forces acting on the body
When the body is released, the two main forces acting on it are:
- The weight of the body, \( mg \), acting downward.
- The air resistance force, \( F_A \), acting upward.

### Step 2: Write the equation of motion
Using Newton's second law, the net force acting on the body can be expressed as:
\[ F_{\text{net}} = mg - F_A = ma \]
where \( a \) is the acceleration of the body.

### Step 3: Determine the acceleration
We can use the kinematic equation to find the acceleration. The equation we will use is:
\[ v^2 = u^2 + 2as \]
Here, \( v = \sqrt{gh} \) (final velocity), \( u = 0 \) (initial velocity), and \( s = h \) (displacement). Plugging in the values, we have:
\[ (\sqrt{gh})^2 = 0 + 2ah \]
This simplifies to:
\[ gh = 2ah \]
Dividing both sides by \( h \) (assuming \( h \neq 0 \)):
\[ g = 2a \implies a = \frac{g}{2} \]

### Step 4: Substitute acceleration back into the equation of motion
Now substituting \( a \) back into the equation of motion:
\[ mg - F_A = m \left(\frac{g}{2}\right) \]
Rearranging gives:
\[ F_A = mg - m\left(\frac{g}{2}\right) = mg - \frac{mg}{2} = \frac{mg}{2} \]

### Step 5: Calculate the work done by air resistance
The work done by the air resistance force \( W_A \) can be calculated using the formula:
\[ W_A = F_A \cdot s \cdot \cos(\theta) \]
where:
- \( F_A = \frac{mg}{2} \)
- \( s = h \)
- \( \theta = 180^\circ \) (since the air resistance acts upward while the displacement is downward)

Thus:
\[ W_A = \frac{mg}{2} \cdot h \cdot \cos(180^\circ) \]
Since \( \cos(180^\circ) = -1 \), we have:
\[ W_A = \frac{mg}{2} \cdot h \cdot (-1) = -\frac{mgh}{2} \]

### Final Answer
The work done by the air resistance force is:
\[ W_A = -\frac{mgh}{2} \]

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