A particle is thrown vertically upwards. If its velocity is half of the maximum height is `20 m//s`, then maximum height attained by it is

Options

A. 35 m
B. 15 m
C. 25 m
D. 40 m

Answer

To solve the problem, we will use the equations of motion. The key points are:

1. The particle is thrown vertically upwards.
2. At the maximum height, the final velocity (v) is 0 m/s.
3. The initial velocity (u) at a certain height is given as half of the maximum height.

Let's denote:
- \( h \) = maximum height attained by the particle
- \( h_1 \) = height at which the velocity is 20 m/s

### Step-by-Step Solution:

1. **Identify the relationship between heights**:
Since the velocity at height \( h_1 \) is given as half of the maximum height, we can write:
\[ h_1 = \frac{h}{2} \]

2. **Use the equation of motion**:
We will use the equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 0 \) m/s (final velocity at maximum height)
- \( u = 20 \) m/s (initial velocity at height \( h_1 \))
- \( a = -g \) (acceleration due to gravity, which is negative as it acts downwards)
- \( s = h_1 \) (the distance traveled to reach maximum height from \( h_1 \))

Plugging in the values:
\[ 0 = (20)^2 + 2(-g)(h_1) \]
This simplifies to:
\[ 0 = 400 - 2gh_1 \]
Rearranging gives:
\[ 2gh_1 = 400 \]
\[ h_1 = \frac{400}{2g} \]

3. **Substituting the value of \( g \)**:
Taking \( g \approx 10 \, \text{m/s}^2 \):
\[ h_1 = \frac{400}{20} = 20 \, \text{m} \]

4. **Finding the maximum height**:
Since \( h_1 = \frac{h}{2} \), we can find \( h \):
\[ h = 2h_1 = 2 \times 20 = 40 \, \text{m} \]

5. **Conclusion**:
The maximum height attained by the particle is:
\[ h = 40 \, \text{m} \]

### Final Answer:
The maximum height attained by the particle is **40 meters**.