A particle is thrown at an angle of `15^(@)` with the horizontal and the range is 1.5 km. What is the range when it is projected at `45^(@)` ?

Options

A. 1.5 km
B. 6 km
C. 4.5 km
D. 3 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 angle of projection,
- \( g \) is the acceleration due to gravity.

### Step 1: Understand the given information
We know that:
- The angle of projection for the first case is \( \theta_1 = 15^\circ \).
- The range for this angle is \( R_1 = 1.5 \) km.
- We need to find the range \( R_2 \) when the angle of projection is \( \theta_2 = 45^\circ \).

### Step 2: Use the range formula
The range for the first angle can be expressed as:

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

And for the second angle:

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

### Step 3: Set up the ratio of the ranges
Since the initial velocity \( u \) and acceleration due to gravity \( g \) remain constant, we can set up the ratio of the two ranges:

\[ \frac{R_1}{R_2} = \frac{\sin(2 \cdot 15^\circ)}{\sin(2 \cdot 45^\circ)} \]

### Step 4: Calculate the sine values
Now, we need to calculate the sine values:

- \( \sin(30^\circ) = \frac{1}{2} \)
- \( \sin(90^\circ) = 1 \)

### Step 5: Substitute the sine values into the ratio
Substituting these values into our ratio gives:

\[ \frac{R_1}{R_2} = \frac{\frac{1}{2}}{1} = \frac{1}{2} \]

### Step 6: Solve for \( R_2 \)
Now, we can rearrange the equation to solve for \( R_2 \):

\[ R_2 = \frac{R_1 \cdot 2}{1} = 2 \cdot R_1 \]

Substituting \( R_1 = 1.5 \) km:

\[ R_2 = 2 \cdot 1.5 \text{ km} = 3 \text{ km} \]

### Final Answer
Thus, the range when the particle is projected at \( 45^\circ \) is:

\[ \boxed{3 \text{ km}} \]