Assume that the acceleration due to gravity on the surface of the moon is `0.2` times the acceleration due to gravity on the surface of the earth. If `R_e` is the maximum range of a projectile on the earth's surface, what is the maximum range on the surface of the moon for the same velocity of projection

Options

A. `0.2 R_(e)`
B. `2R_(e)`
C. `0.5 R_(e)`
D. `5R_(e)`

Explanation

To solve the problem, we need to find the maximum range of a projectile on the surface of the Moon, given that the acceleration due to gravity on the Moon is \(0.2\) times that on Earth. We will start by recalling the formula for the range of a projectile.

### Step-by-Step Solution:

1. **Understand the Range Formula**:
The range \(R\) of a projectile launched with an initial velocity \(U\) at an angle \(\theta\) is given by:
\[ R = \frac{U^2 \sin 2\theta}{g} \]
where \(g\) is the acceleration due to gravity.

2. **Determine Maximum Range on Earth**:
The maximum range occurs when \(\theta = 45^\circ\) (since \(\sin 90^\circ = 1\)). Thus, the maximum range on Earth \(R_e\) can be expressed as:
\[ R_e = \frac{U^2}{g_E} \]
where \(g_E\) is the acceleration due to gravity on Earth.

3. **Calculate the Range on the Moon**:
The acceleration due to gravity on the Moon \(g_M\) is given as:
\[ g_M = 0.2 \times g_E \]
The maximum range on the Moon \(R_m\) can be expressed similarly:
\[ R_m = \frac{U^2}{g_M} \]

4. **Substituting for \(g_M\)**:
Substitute \(g_M\) into the equation for \(R_m\):
\[ R_m = \frac{U^2}{0.2 \times g_E} \]

5. **Relate \(R_m\) to \(R_e\)**:
Now, we can express \(R_m\) in terms of \(R_e\):
\[ R_m = \frac{U^2}{0.2 \times g_E} = \frac{1}{0.2} \cdot \frac{U^2}{g_E} = 5 \cdot R_e \]

6. **Final Result**:
Therefore, the maximum range on the Moon is:
\[ R_m = 5 \cdot R_e \]

### Conclusion:
The maximum range of the projectile on the surface of the Moon is \(5\) times the maximum range on the surface of the Earth.