A stone of weight W is thrown straight up from the ground with an intial speed V0. If a drag force of constant magnitude f acts - on the same through out its flight , the speed of stone just before reaching the ground is
(1) `V0sqrt((W-f)/W)` (2) `V0sqrt((W+f)/f)`
(3)` V0 sqrt((W+f)/(W-f))` (4) ` V0 sqrt((W-f)/(W+f))`

Answer

To solve the problem, we need to determine the speed of a stone just before it reaches the ground after being thrown upwards with an initial speed \( V_0 \) and experiencing a constant drag force \( f \) throughout its flight.

### Step-by-Step Solution:

1. **Identify the forces acting on the stone:**
- The weight of the stone \( W \) acts downward.
- The drag force \( f \) also acts downward when the stone is moving upwards and acts upward when the stone is falling down.

2. **Determine the acceleration during the upward motion:**
- When the stone is thrown upwards, the net force acting on it is:
\[ F_{\text{net, up}} = W + f \]
- The acceleration \( a \) during the upward motion is given by:
\[ a = \frac{F_{\text{net, up}}}{m} = \frac{W + f}{m} \]

3. **Use the kinematic equation for upward motion:**
- The kinematic equation we can use is:
\[ V_f^2 = V_0^2 - 2aS \]
- Here, \( V_f \) is the final velocity at the peak (which is 0), \( V_0 \) is the initial velocity, \( a \) is the acceleration, and \( S \) is the maximum height reached.
- Rearranging gives:
\[ 0 = V_0^2 - 2\left(\frac{W + f}{m}\right)S \]
- Solving for \( S \):
\[ S = \frac{V_0^2 m}{2(W + f)} \]

4. **Determine the acceleration during the downward motion:**
- When the stone falls back down, the net force acting on it is:
\[ F_{\text{net, down}} = W - f \]
- The acceleration \( a' \) during the downward motion is:
\[ a' = \frac{F_{\text{net, down}}}{m} = \frac{W - f}{m} \]

5. **Use the kinematic equation for downward motion:**
- The kinematic equation we can use is:
\[ V_f^2 = V_i^2 + 2a'S \]
- Here, \( V_i \) is the initial velocity just before falling (which is 0 at the peak), and \( S \) is the distance fallen (which is the same as the height reached).
- Substituting gives:
\[ V_f^2 = 0 + 2\left(\frac{W - f}{m}\right)S \]
- Substituting for \( S \) from step 3:
\[ V_f^2 = 2\left(\frac{W - f}{m}\right)\left(\frac{V_0^2 m}{2(W + f)}\right) \]
- Simplifying this gives:
\[ V_f^2 = \frac{(W - f)V_0^2}{W + f} \]

6. **Taking the square root to find \( V_f \):**
- Finally, we find:
\[ V_f = V_0 \sqrt{\frac{W - f}{W + f}} \]

### Final Answer:
The speed of the stone just before reaching the ground is:
\[ V_f = V_0 \sqrt{\frac{W - f}{W + f}} \]