A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to 1/3rd the escape velocity for the planet. The maximum height attained by the body is

Options

A. `R//2`
B. `R//3`
C. `R//5`
D. `R//9`

Explanation

To solve the problem, we need to find the maximum height attained by a body projected vertically upwards from the surface of a planet with a velocity equal to one-third of the escape velocity.

### Step-by-Step Solution:

1. **Understand the Escape Velocity**:
The escape velocity \( v_e \) from the surface of a planet is given by the formula:
\[ v_e = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity at the surface of the planet and \( R \) is the radius of the planet.

2. **Determine the Initial Velocity**:
According to the problem, the body is projected with a velocity equal to one-third of the escape velocity:
\[ u = \frac{1}{3} v_e = \frac{1}{3} \sqrt{2gR} \]

3. **Use the Equation of Motion**:
The maximum height \( h \) can be found using the kinematic equation:
\[ v^2 = u^2 + 2(-g)h \]
At the maximum height, the final velocity \( v \) becomes 0. Thus, the equation simplifies to:
\[ 0 = u^2 - 2gh \]
Rearranging gives:
\[ u^2 = 2gh \]

4. **Substituting the Initial Velocity**:
Substitute the expression for \( u \) into the equation:
\[ \left(\frac{1}{3} \sqrt{2gR}\right)^2 = 2gh \]
Simplifying the left side:
\[ \frac{1}{9} (2gR) = 2gh \]
This leads to:
\[ \frac{2gR}{9} = 2gh \]

5. **Solve for Maximum Height \( h \)**:
Dividing both sides by \( 2g \):
\[ \frac{R}{9} = h \]
Therefore, the maximum height attained by the body is:
\[ h = \frac{R}{9} \]

### Final Answer:
The maximum height attained by the body is \( \frac{R}{9} \).

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