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. `R//6`
B. `6R`
C. `R//36`
D. `36R`

Answer

To solve the problem of finding the range \( R' \) of a projectile thrown with the same speed \( v \) and angle \( \theta \) on the surface of the Moon, we can use the formula for the range of a projectile:

\[ R = \frac{v^2 \sin 2\theta}{g} \]

where:
- \( R \) is the range,
- \( v \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step 1: Determine the range on Earth

For the Earth, the range \( R \) can be expressed as:

\[ R = \frac{v^2 \sin 2\theta}{g_{earth}} \]

where \( g_{earth} \) is approximately \( 9.8 \, \text{m/s}^2 \).

### Step 2: Determine the range on the Moon

For the Moon, the range \( R' \) can be expressed similarly:

\[ R' = \frac{v^2 \sin 2\theta}{g_{moon}} \]

where \( g_{moon} \) is approximately \( 1.6 \, \text{m/s}^2 \).

### Step 3: Find the ratio of the ranges

To find the relationship between the range on the Moon and the range on Earth, we can take the ratio:

\[ \frac{R'}{R} = \frac{\frac{v^2 \sin 2\theta}{g_{moon}}}{\frac{v^2 \sin 2\theta}{g_{earth}}} \]

This simplifies to:

\[ \frac{R'}{R} = \frac{g_{earth}}{g_{moon}} \]

### Step 4: Substitute the values of \( g \)

Substituting the values of \( g_{earth} \) and \( g_{moon} \):

\[ \frac{R'}{R} = \frac{9.8}{1.6} \]

Calculating this gives:

\[ \frac{R'}{R} \approx 6.125 \]

### Step 5: Calculate \( R' \)

Thus, we can express \( R' \) in terms of \( R \):

\[ R' = 6.125 R \]

### Conclusion

Therefore, the range of the projectile on the surface of the Moon is approximately \( 6.125 \) times the range on the surface of the Earth.

### Final Answer

The range \( R' \) on the surface of the Moon will be approximately \( 6R \).

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