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 use the formula for the range of a projectile, which is given by:

\[ R = \frac{u^2 \sin(2\theta)}{g} \]

where:
- \( R \) is the range,
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 1: Calculate \( \frac{u^2}{g} \) using the first angle (15 degrees)

Given that the range \( R \) when the angle \( \theta = 15^\circ \) is 50 m, we can substitute the values into the range formula:

\[ 50 = \frac{u^2 \sin(30^\circ)}{g} \]

Since \( \sin(30^\circ) = \frac{1}{2} \), we can rewrite the equation as:

\[ 50 = \frac{u^2 \cdot \frac{1}{2}}{g} \]

Multiplying both sides by \( 2g \):

\[ 100g = u^2 \]

Thus, we find:

\[ \frac{u^2}{g} = 100 \]

### Step 2: Calculate the range for the second angle (45 degrees)

Now we need to find the range when the angle \( \theta = 45^\circ \):

\[ R = \frac{u^2 \sin(90^\circ)}{g} \]

Since \( \sin(90^\circ) = 1 \), we can substitute \( \frac{u^2}{g} = 100 \):

\[ R = \frac{100g \cdot 1}{g} \]

The \( g \) cancels out:

\[ R = 100 \, \text{m} \]

### Conclusion

The range of the projectile when fired at an angle of \( 45^\circ \) is **100 meters**.

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