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 need to find the range of a projectile when it is fired at an angle of \(45^\circ\), given that its range at an angle of \(15^\circ\) is \(50\) m.

### Step-by-Step Solution:

1. **Understand the Range Formula**:
The range \(R\) of a projectile 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.

2. **Set Up the Equations**:
For the first projectile fired at an angle of \(15^\circ\):
\[ R_1 = \frac{u^2 \sin(2 \times 15^\circ)}{g} \]
For the second projectile fired at an angle of \(45^\circ\):
\[ R_2 = \frac{u^2 \sin(2 \times 45^\circ)}{g} \]

3. **Calculate the Sine Values**:
- \(\sin(30^\circ) = \frac{1}{2}\)
- \(\sin(90^\circ) = 1\)

4. **Express the Ratios**:
Now, we can set up the ratio of the ranges:
\[ \frac{R_1}{R_2} = \frac{\sin(30^\circ)}{\sin(90^\circ)} \]

5. **Substituting Known Values**:
We know that \(R_1 = 50\) m, so:
\[ \frac{50}{R_2} = \frac{\frac{1}{2}}{1} \]

6. **Cross-Multiply to Solve for \(R_2\)**:
\[ 50 = \frac{1}{2} R_2 \]
Multiplying both sides by \(2\):
\[ R_2 = 100 \text{ m} \]

### Final Answer:
The range of the projectile when fired at an angle of \(45^\circ\) is \(100\) m.

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