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. 25m
B. 37m
C. 50m
D. 100m

Answer

To solve the problem, we will 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.

### Step 1: Calculate \( \frac{u^2}{g} \) using the first range
We know that when the projectile is fired at an angle of \( 15^\circ \), the range \( R_1 \) is 50 m.

Using the formula:
\[ R_1 = \frac{u^2 \sin(2 \cdot 15^\circ)}{g} \]

Calculating \( \sin(30^\circ) \):
\[ \sin(30^\circ) = \frac{1}{2} \]

Substituting the values:
\[ 50 = \frac{u^2 \cdot \frac{1}{2}}{g} \]

Rearranging gives:
\[ 50g = \frac{u^2}{2} \]
\[ u^2 = 100g \]

### Step 2: Calculate the range for the angle of \( 45^\circ \)
Now, we need to find the range when the projectile is fired at an angle of \( 45^\circ \).

Using the same formula:
\[ R_2 = \frac{u^2 \sin(2 \cdot 45^\circ)}{g} \]

Calculating \( \sin(90^\circ) \):
\[ \sin(90^\circ) = 1 \]

Substituting the values we found:
\[ R_2 = \frac{u^2 \cdot 1}{g} \]
\[ R_2 = \frac{100g}{g} \]
\[ R_2 = 100 \, \text{m} \]

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

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