The acceleration due to gravity on the moon is 1//6th of that on the earth. If a man can jump upto a height of 1 m on the surface of the earth, how high he can jump on the surface of the moon ?

Options

A. 3 m
B. 4 m
C. 6 m
D. 8 m

Explanation

To solve the problem, we need to understand how the height a person can jump is affected by the acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the given information**:
- The height a man can jump on Earth, \( h_{E} = 1 \, \text{m} \).
- The acceleration due to gravity on the Moon, \( g_{M} = \frac{1}{6} g_{E} \), where \( g_{E} \) is the acceleration due to gravity on Earth.

2. **Use the formula for potential energy**:
- The potential energy (PE) at the maximum height reached during a jump can be expressed as:
\[ PE = mgh \]
- On Earth, the potential energy when jumping to a height of 1 m is:
\[ PE_{E} = mg_{E} \cdot h_{E} = mg_{E} \cdot 1 \]

3. **Set up the equation for the Moon**:
- On the Moon, the potential energy at the maximum height \( h_{M} \) can be expressed as:
\[ PE_{M} = mg_{M} \cdot h_{M} \]
- Since \( g_{M} = \frac{1}{6} g_{E} \), we can substitute this into the equation:
\[ PE_{M} = mg_{M} \cdot h_{M} = m \left(\frac{1}{6} g_{E}\right) \cdot h_{M} \]

4. **Equate the potential energies**:
- Since the man uses the same amount of energy to jump on both the Earth and the Moon, we can set the two potential energies equal to each other:
\[ mg_{E} \cdot 1 = m \left(\frac{1}{6} g_{E}\right) \cdot h_{M} \]

5. **Cancel out the mass and gravitational acceleration**:
- The mass \( m \) cancels out from both sides:
\[ g_{E} = \frac{1}{6} g_{E} \cdot h_{M} \]
- Now, divide both sides by \( g_{E} \):
\[ 1 = \frac{1}{6} h_{M} \]

6. **Solve for \( h_{M} \)**:
- Multiply both sides by 6 to isolate \( h_{M} \):
\[ h_{M} = 6 \, \text{m} \]

### Conclusion:
The height a man can jump on the surface of the Moon is **6 meters**.