The range of particle when launched at an angle of `15^(@)` with the horizontal is 1.5 km. What is the range of the projectile when launched at an angle of `45^(@)` to the horizontal

Options

A. 1.5 km
B. 3.0 km
C. 6.0 km
D. 0.75 km

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 launch angle,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) or \( 10 \, \text{m/s}^2 \) for simplicity).

### Step 1: Calculate the initial velocity \( u \) using the given range at \( 15^\circ \)

Given:
- Range \( R_1 = 1.5 \, \text{km} = 1500 \, \text{m} \)
- Launch angle \( \theta_1 = 15^\circ \)

Using the range formula:

\[ R_1 = \frac{u^2 \sin(2 \theta_1)}{g} \]

Substituting the known values:

\[ 1500 = \frac{u^2 \sin(30^\circ)}{10} \]

Since \( \sin(30^\circ) = \frac{1}{2} \):

\[ 1500 = \frac{u^2 \cdot \frac{1}{2}}{10} \]

This simplifies to:

\[ 1500 = \frac{u^2}{20} \]

Multiplying both sides by 20:

\[ u^2 = 1500 \times 20 = 30000 \]

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

Now we need to find the range when the projectile is launched at an angle of \( 45^\circ \):

Given:
- Launch angle \( \theta_2 = 45^\circ \)

Using the range formula again:

\[ R_2 = \frac{u^2 \sin(2 \theta_2)}{g} \]

Substituting the known values:

\[ R_2 = \frac{u^2 \sin(90^\circ)}{10} \]

Since \( \sin(90^\circ) = 1 \):

\[ R_2 = \frac{u^2}{10} \]

Now substituting \( u^2 = 30000 \):

\[ R_2 = \frac{30000}{10} = 3000 \, \text{m} = 3 \, \text{km} \]

### Final Answer

The range of the projectile when launched at an angle of \( 45^\circ \) is **3 km**.

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