A body is thrown vertically up with certain velocity. If h is the maximum height reached by it, its position when its velocity reduces to 1/3 of its velocity of projection is at

Options

A. `(8h)/9` from the ground
B. `(8h)/9` from the top most point
C. `(4h)/9` from the ground
D. `h/3` from the top most point

Answer

To solve the problem step by step, we will follow the physics of motion under gravity, specifically for a body thrown vertically upwards.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A body is thrown vertically upwards with an initial velocity \( u \).
- The maximum height reached by the body is denoted as \( h \).
- We need to find the position of the body when its velocity reduces to \( \frac{1}{3} \) of the initial velocity \( u \).

2. **Maximum Height Calculation**:
- The maximum height \( h \) can be calculated using the formula:
\[ h = \frac{u^2}{2g} \]
- Here, \( g \) is the acceleration due to gravity.

3. **Velocity at a Certain Height**:
- When the velocity of the body reduces to \( \frac{u}{3} \), we can denote this velocity as \( v = \frac{u}{3} \).

4. **Using the Third Equation of Motion**:
- We will use the third equation of motion:
\[ v^2 = u^2 + 2as \]
- In our case, the acceleration \( a \) is \( -g \) (since gravity acts downwards), and \( s \) is the height we need to find.
- Substituting the values, we have:
\[ \left(\frac{u}{3}\right)^2 = u^2 - 2gs \]

5. **Simplifying the Equation**:
- Expanding the left side:
\[ \frac{u^2}{9} = u^2 - 2gs \]
- Rearranging gives:
\[ 2gs = u^2 - \frac{u^2}{9} \]
- Finding a common denominator on the right side:
\[ 2gs = \frac{9u^2}{9} - \frac{u^2}{9} = \frac{8u^2}{9} \]

6. **Solving for \( s \)**:
- Now, dividing both sides by \( 2g \):
\[ s = \frac{8u^2}{18g} = \frac{4u^2}{9g} \]

7. **Relating \( s \) to \( h \)**:
- Recall that \( h = \frac{u^2}{2g} \).
- We can express \( s \) in terms of \( h \):
\[ s = \frac{4u^2}{9g} = \frac{4}{9} \cdot \frac{u^2}{2g} = \frac{4}{9}h \]

8. **Finding the Position from the Ground**:
- Since the body reaches a maximum height \( h \), the position when the velocity is \( \frac{u}{3} \) is:
\[ h - s = h - \frac{4}{9}h = \frac{5}{9}h \]
- Therefore, the position of the body when its velocity is \( \frac{1}{3} \) of the initial velocity is at \( \frac{5}{9}h \) from the ground.

### Final Answer:
The position of the body when its velocity reduces to \( \frac{1}{3} \) of its initial velocity is at \( \frac{5h}{9} \) from the ground.