The range of a projectile, 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 with the same speed ?

Options

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

Answer

To solve the problem, we will use the formula for the range of a projectile, which is given by:

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

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 \( V^2/g \) using the given range at \( 15^\circ \)

Given that the range \( R_1 \) when launched at \( 15^\circ \) is \( 1.5 \, \text{km} \) (or \( 1500 \, \text{m} \)), we can write:

\[ R_1 = \frac{V^2 \sin(30^\circ)}{g} \]

Since \( \sin(30^\circ) = \frac{1}{2} \), we can substitute this into the equation:

\[ 1500 = \frac{V^2 \cdot \frac{1}{2}}{g} \]

Rearranging gives us:

\[ V^2 = \frac{1500 \cdot 2g}{1} = 3000g \]

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

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

\[ R_2 = \frac{V^2 \sin(90^\circ)}{g} \]

Since \( \sin(90^\circ) = 1 \), we can substitute \( V^2 \) from Step 1:

\[ R_2 = \frac{3000g \cdot 1}{g} = 3000 \, \text{m} \]

### Final Result

Converting \( R_2 \) back to kilometers:

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

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

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