The acceleration due to gravity on the planet `A` is `9` times the acceleration due to gravity on planet `B`. A man jumps to a height of `2 m` on the surface of `A`. What is the height of jump by the same person on the planet `B` ?

Options

A. 6m
B. `(2/3)` m
C. `(2/9)` m
D. 18 m

Explanation

To solve the problem, we need to find the height a man can jump on planet B given that he jumps to a height of 2 meters on planet A, where the acceleration due to gravity on planet A is 9 times that on planet B.

### Step-by-Step Solution:

1. **Identify the given values:**
- Let \( g_A \) be the acceleration due to gravity on planet A.
- Let \( g_B \) be the acceleration due to gravity on planet B.
- We know that \( g_A = 9g_B \).
- The height jumped on planet A, \( H_A = 2 \, \text{m} \).

2. **Use the formula for maximum height in terms of initial velocity and gravity:**
The maximum height \( H \) reached by a person when jumping can be expressed as:
\[ H = \frac{v^2}{2g} \]
where \( v \) is the initial velocity of the jump.

3. **Set up the equation for height on planet A:**
For planet A:
\[ H_A = \frac{v^2}{2g_A} \]
Substituting the known height:
\[ 2 = \frac{v^2}{2g_A} \]

4. **Rearranging the equation to find \( v^2 \):**
Multiplying both sides by \( 2g_A \):
\[ v^2 = 4g_A \]

5. **Set up the equation for height on planet B:**
For planet B:
\[ H_B = \frac{v^2}{2g_B} \]

6. **Substituting \( v^2 \) from planet A into the equation for planet B:**
\[ H_B = \frac{4g_A}{2g_B} \]
Simplifying this gives:
\[ H_B = \frac{2g_A}{g_B} \]

7. **Substituting \( g_A = 9g_B \) into the equation for \( H_B \):**
\[ H_B = \frac{2(9g_B)}{g_B} \]
The \( g_B \) cancels out:
\[ H_B = 2 \times 9 = 18 \, \text{m} \]

### Final Answer:
The height of the jump by the same person on planet B is \( 18 \, \text{m} \).