The range of a projectile fired at an angle of `15^@` is 50 m. If it is fired with the same speed at an angle of `45^@` its range will be

Options

A. 60 m
B. 71 m
C. 100 m
D. 141 m

Explanation

To solve the problem, we need to find the range of a projectile fired at an angle of \( 45^\circ \) given that its range at an angle of \( 15^\circ \) is \( 50 \, \text{m} \).

### Step-by-Step Solution:

1. **Recall the formula for the range of a projectile:**
The range \( R \) of a projectile launched with an initial speed \( u \) at an angle \( \theta \) is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

2. **Calculate the range for \( 15^\circ \):**
Given that the range at \( 15^\circ \) is \( 50 \, \text{m} \), we can write:
\[ 50 = \frac{u^2 \sin(30^\circ)}{g} \]
Since \( \sin(30^\circ) = \frac{1}{2} \), we can substitute this into the equation:
\[ 50 = \frac{u^2 \cdot \frac{1}{2}}{g} \]

3. **Rearranging the equation:**
Multiplying both sides by \( g \) gives:
\[ 50g = \frac{u^2}{2} \]
Now, multiplying both sides by \( 2 \):
\[ 100g = u^2 \]

4. **Calculate the range for \( 45^\circ \):**
Now, we need to find the range when the projectile is fired at \( 45^\circ \):
\[ R = \frac{u^2 \sin(90^\circ)}{g} \]
Since \( \sin(90^\circ) = 1 \), we can simplify this to:
\[ R = \frac{u^2}{g} \]

5. **Substituting \( u^2 \) from the previous calculation:**
We already found \( u^2 = 100g \). Substituting this into the range formula gives:
\[ R = \frac{100g}{g} \]
Simplifying this results in:
\[ R = 100 \, \text{m} \]

### Conclusion:
The range of the projectile when fired at an angle of \( 45^\circ \) is \( 100 \, \text{m} \).