A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s, then maximum height attained by it is

Options

A. 8m
B. 10m
C. 12m
D. 16m

Explanation

To solve the problem, we need to find the maximum height attained by a particle thrown vertically upwards, given that its velocity at half of the maximum height is 10 m/s.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We know that the particle is thrown upwards and reaches a maximum height \( H \).
- At half of this height (\( \frac{H}{2} \)), the velocity \( V \) is given as 10 m/s.

2. **Using the Equation of Motion**:
- We can use the third equation of motion, which is:
\[ V^2 = U^2 + 2a s \]
- Here, \( V \) is the final velocity (10 m/s), \( U \) is the initial velocity (unknown), \( a \) is the acceleration (which will be \(-g\) or \(-10 \, \text{m/s}^2\) since gravity acts downwards), and \( s \) is the displacement (\( \frac{H}{2} \)).
- Rearranging the equation gives:
\[ 10^2 = U^2 - 2g \left(\frac{H}{2}\right) \]
- This simplifies to:
\[ 100 = U^2 - 10H \quad \text{(Equation 1)} \]

3. **Finding the Initial Velocity \( U \)**:
- Now we need another equation to express \( U \) in terms of \( H \). We will use the same equation of motion for the entire height \( H \):
\[ 0 = U^2 - 2gH \]
- Rearranging gives:
\[ U^2 = 2gH \quad \text{(Equation 2)} \]

4. **Substituting Equation 2 into Equation 1**:
- Substitute \( U^2 \) from Equation 2 into Equation 1:
\[ 100 = 2gH - 10H \]
- Substituting \( g = 10 \, \text{m/s}^2 \):
\[ 100 = 20H - 10H \]
- This simplifies to:
\[ 100 = 10H \]

5. **Solving for Maximum Height \( H \)**:
- Dividing both sides by 10 gives:
\[ H = 10 \, \text{meters} \]

### Conclusion:
The maximum height attained by the particle is **10 meters**.