A stone with weight W is thrown vertically upward into the air with initial velocity `v_(0)`. If a constant forcef due to air drag acts on the stone throughout the flight & if the maximum height attain by stone is h and velocity when it strikes to the ground is v. Which one is correct?

Options

A. `h-(v_(0)^(2))/(2g(1+(f)/(w)))`
B. `h=(v_(0)^(2))/(2g(1-(f)/(w)))`
C. `h=(v_(0)^(2))/(2g(1+(w)/(f)))`
D. `h=(v_(0)^(2))/(2g(1-(w)/(f)))`

Answer

To solve the problem step by step, we will use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

### Step 1: Understand the scenario
A stone is thrown vertically upward with an initial velocity \( v_0 \). It experiences two forces during its flight: the gravitational force \( W \) (weight of the stone) acting downward and a constant air drag force \( f \) also acting downward.

### Step 2: Identify the maximum height
At the maximum height \( h \), the final velocity of the stone is zero. Therefore, we can apply the work-energy theorem from the point of release to the maximum height.

### Step 3: Write the work-energy theorem
The work-energy theorem can be expressed as:
\[ \Delta KE = W_{net} \]
Where:
- \( \Delta KE \) is the change in kinetic energy.
- \( W_{net} \) is the net work done on the object.

### Step 4: Calculate the change in kinetic energy
The initial kinetic energy when the stone is thrown is:
\[ KE_{initial} = \frac{1}{2} m v_0^2 \]
The final kinetic energy at the maximum height is:
\[ KE_{final} = 0 \]
Thus, the change in kinetic energy is:
\[ \Delta KE = KE_{final} - KE_{initial} = 0 - \frac{1}{2} m v_0^2 = -\frac{1}{2} m v_0^2 \]

### Step 5: Calculate the work done by forces
The work done against gravity (weight) is:
\[ W_{gravity} = -mgh \]
The work done against air drag is:
\[ W_{drag} = -fh \]
Thus, the total work done \( W_{net} \) is:
\[ W_{net} = W_{gravity} + W_{drag} = -mgh - fh \]

### Step 6: Set up the equation
Now, we can equate the change in kinetic energy to the net work done:
\[ -\frac{1}{2} m v_0^2 = -mgh - fh \]

### Step 7: Rearrange the equation
Rearranging gives us:
\[ mgh + fh = \frac{1}{2} m v_0^2 \]
Factoring out \( h \):
\[ h(mg + f) = \frac{1}{2} m v_0^2 \]

### Step 8: Solve for maximum height \( h \)
Now, we can solve for \( h \):
\[ h = \frac{\frac{1}{2} m v_0^2}{mg + f} \]

### Step 9: Substitute \( m \) with \( W \)
Since \( W = mg \), we can replace \( m \) in the equation:
\[ h = \frac{\frac{1}{2} \frac{W}{g} v_0^2}{W + f} \]
This simplifies to:
\[ h = \frac{W v_0^2}{2g(W + f)} \]

### Step 10: Final expression
Thus, the final expression for the maximum height \( h \) is:
\[ h = \frac{v_0^2}{2g} \left(1 + \frac{f}{W}\right) \]

### Conclusion
The correct option is:
\[ \frac{v_0^2}{2g} \left(1 + \frac{f}{W}\right) \]