The range of a projectile when fired at `75^(@)` with the horizontal is 0.5km. What will be its range when fired at `45^(@)` with same speed:-

Options

A. 0.5 km
B. 1.0 km
C. 1.5 km
D. 2.0 km

Answer

To solve the problem, we will use the formula for the range of a projectile:

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

where:
- \( R \) is the range,
- \( V \) 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 the initial velocity \( V \) using the range at \( 75^\circ \)

Given that the range \( R_1 \) when fired at \( 75^\circ \) is \( 0.5 \, \text{km} = 500 \, \text{m} \):

\[ R_1 = \frac{V^2 \sin(2 \times 75^\circ)}{g} \]

Calculating \( \sin(150^\circ) \):

\[ \sin(150^\circ) = \sin(30^\circ) = \frac{1}{2} \]

Substituting into the range formula:

\[ 500 = \frac{V^2 \cdot \frac{1}{2}}{g} \]

Rearranging gives:

\[ V^2 = \frac{500 \cdot 2g}{1} = 1000g \]

### Step 2: Calculate the range \( R_2 \) when fired at \( 45^\circ \)

Now, we need to find the range \( R_2 \) when the projectile is fired at \( 45^\circ \):

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

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

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

Substituting \( V^2 \) from the previous step:

\[ R_2 = \frac{1000g \cdot 1}{g} \]

The \( g \) cancels out:

\[ R_2 = 1000 \, \text{m} = 1 \, \text{km} \]

### Conclusion

The range when fired at \( 45^\circ \) with the same speed is \( 1 \, \text{km} \).

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