A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is `10 ms^-1` , then the maximum height attained by the stone is (`g= 10 ms^(-2)`)

Options

A. 25 m
B. 10 m
C. 15 m
D. 20 m

Explanation

To solve the problem, we will use the kinematic equations of motion. The key points are that the stone is thrown vertically upwards, and we know its speed at half of its maximum height.

### Step-by-Step Solution:

1. **Identify Known Values**:
- Speed at half of maximum height, \( v = 10 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Define Maximum Height**:
- Let the maximum height attained by the stone be \( H \).
- Therefore, half of the maximum height is \( \frac{H}{2} \).

3. **Use the Kinematic Equation**:
- At the height \( \frac{H}{2} \), we can use the kinematic equation:
\[ v^2 = u^2 - 2g\left(\frac{H}{2}\right) \]
- Here, \( u \) is the initial velocity (which we need to find), and \( v \) is the final velocity at height \( \frac{H}{2} \).

4. **Substitute Known Values**:
- Substitute \( v = 10 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ 10^2 = u^2 - 2 \cdot 10 \cdot \left(\frac{H}{2}\right) \]
- This simplifies to:
\[ 100 = u^2 - 10H \]
- Rearranging gives:
\[ u^2 = 100 + 10H \quad \text{(Equation 1)} \]

5. **Use Another Kinematic Equation**:
- At the maximum height \( H \), the final velocity \( v = 0 \):
\[ 0 = u^2 - 2gH \]
- Rearranging gives:
\[ u^2 = 2gH \]
- Substitute \( g = 10 \, \text{m/s}^2 \):
\[ u^2 = 20H \quad \text{(Equation 2)} \]

6. **Equate the Two Expressions for \( u^2 \)**:
- From Equation 1 and Equation 2:
\[ 100 + 10H = 20H \]
- Rearranging gives:
\[ 100 = 20H - 10H \]
\[ 100 = 10H \]
- Therefore:
\[ H = 10 \, \text{m} \]

### Final Answer:
The maximum height attained by the stone is \( H = 10 \, \text{m} \).

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