A projectile thrown with a speed v at an angle `theta` has a range R on the surface of earth. For same v and `theta`, its range on the surface of moon will be

Options

A. 36R
B. `(R)/(36)`
C. `(R)/(16)`
D. 6R

Explanation

To find the range of a projectile thrown with a speed \( v \) at an angle \( \theta \) on the surface of the Moon, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Range Formula on Earth**:
The range \( R \) of a projectile on Earth is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity on Earth.

2. **Substituting Known Values**:
For the projectile thrown with speed \( v \) at angle \( \theta \), we can substitute \( u = v \):
\[ R = \frac{v^2 \sin 2\theta}{g} \]

3. **Gravitational Acceleration on the Moon**:
The gravitational acceleration on the Moon is approximately \( \frac{1}{6} \)th of that on Earth:
\[ g_{moon} = \frac{g}{6} \]

4. **Range Formula on the Moon**:
The range \( R_{moon} \) on the Moon can be calculated using the same range formula, but substituting \( g_{moon} \):
\[ R_{moon} = \frac{u^2 \sin 2\theta}{g_{moon}} \]
Substituting \( u = v \) and \( g_{moon} = \frac{g}{6} \):
\[ R_{moon} = \frac{v^2 \sin 2\theta}{\frac{g}{6}} = 6 \cdot \frac{v^2 \sin 2\theta}{g} \]

5. **Relating to Earth’s Range**:
From our earlier calculation, we know that:
\[ R = \frac{v^2 \sin 2\theta}{g} \]
Therefore, we can express \( R_{moon} \) in terms of \( R \):
\[ R_{moon} = 6R \]

### Final Answer:
The range of the projectile on the surface of the Moon is:
\[ R_{moon} = 6R \]