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

Explanation

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 (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 1: Calculate the initial velocity \( u \) using the first range
Given that the range \( R_1 \) at an angle of \( 15^\circ \) is \( 50 \, \text{m} \):

\[ R_1 = 50 = \frac{u^2 \sin(2 \times 15^\circ)}{g} \]

First, we need to calculate \( \sin(30^\circ) \):

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

Now substituting the values into the range equation:

\[ 50 = \frac{u^2 \cdot \frac{1}{2}}{9.8} \]

### Step 2: Rearranging to find \( u^2 \)
Multiplying both sides by \( 9.8 \):

\[ 50 \cdot 9.8 = \frac{u^2}{2} \]

\[ 490 = \frac{u^2}{2} \]

Now, multiplying both sides by \( 2 \):

\[ u^2 = 980 \]

### Step 3: Calculate the range at \( 45^\circ \)
Now we need to find the range \( R_2 \) when the angle is \( 45^\circ \):

\[ R_2 = \frac{u^2 \sin(2 \times 45^\circ)}{g} \]

Calculating \( \sin(90^\circ) \):

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

Substituting the values into the range equation:

\[ R_2 = \frac{980 \cdot 1}{9.8} \]

### Step 4: Simplifying to find \( R_2 \)
Now, simplifying the expression:

\[ R_2 = \frac{980}{9.8} = 100 \, \text{m} \]

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

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

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