A balloon is moving upwards with velocity `10m//s`. It releases a stone which comes down to the ground in `11s`. The height of the balloon from the ground at the moment when the stone was dropped is `(g=10m//s^(2))`

Options

A. `495m`
B. `592m`
C. `362m`
D. `500m`

Answer

To solve the problem, we need to find the height of the balloon from the ground at the moment when the stone was dropped. We will use the equations of motion to do this.

### Step-by-Step Solution:

1. **Identify the given values:**
- Velocity of the balloon (upward) \( V = 10 \, \text{m/s} \)
- Time taken by the stone to reach the ground \( t = 11 \, \text{s} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

2. **Determine the initial velocity of the stone:**
- When the stone is released from the balloon, it has the same initial velocity as the balloon but in the downward direction. Therefore, we take the initial velocity of the stone as:
\[ V = -10 \, \text{m/s} \]
(negative because it is moving downward).

3. **Use the equation of motion to find the height \( H \):**
- The equation of motion we will use is:
\[ H = V t + \frac{1}{2} a t^2 \]
- Here, \( a \) is the acceleration due to gravity, which is \( g = 10 \, \text{m/s}^2 \).

4. **Substituting the values into the equation:**
- Substitute \( V = -10 \, \text{m/s} \), \( a = 10 \, \text{m/s}^2 \), and \( t = 11 \, \text{s} \):
\[ H = (-10) \cdot (11) + \frac{1}{2} \cdot (10) \cdot (11^2) \]

5. **Calculating each term:**
- First term:
\[ -10 \cdot 11 = -110 \]
- Second term:
\[ \frac{1}{2} \cdot 10 \cdot 121 = 5 \cdot 121 = 605 \]

6. **Combine the results:**
- Now, combine the two terms:
\[ H = -110 + 605 = 495 \, \text{m} \]

7. **Conclusion:**
- The height of the balloon from the ground at the moment when the stone was dropped is:
\[ \boxed{495 \, \text{m}} \]