The horizontal range of an oblique projectile is equal to the distance through which a projectile has to fall freely from rest to acquire a velocity equal to the velocity of projection in magnitude. The angle of projection is

Options

A. `75^@`
B. `60^@`
C. `45^@`
D. `30^@`

Answer

To solve the problem, we need to find the angle of projection (θ) such that the horizontal range (R) of an oblique projectile is equal to the distance (h) through which a projectile has to fall freely from rest to acquire a velocity equal to the velocity of projection (u) in magnitude.

### Step-by-Step Solution:

1. **Understand the Range of Projectile Motion**:
The formula for the horizontal range \( R \) of a projectile launched with an initial velocity \( u \) at an angle \( \theta \) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( g \) is the acceleration due to gravity.

2. **Free Fall Distance**:
When an object falls freely from rest, the distance \( h \) it falls to reach a velocity \( u \) can be derived from the third equation of motion:
\[ v^2 = u^2 + 2gh \]
Here, the initial velocity \( u = 0 \) (since it falls from rest), and \( v \) becomes \( u \) (the final velocity). Thus, we have:
\[ u^2 = 0 + 2gh \implies h = \frac{u^2}{2g} \]

3. **Equate the Two Distances**:
According to the problem, the horizontal range \( R \) is equal to the distance \( h \):
\[ R = h \]
Substituting the expressions for \( R \) and \( h \):
\[ \frac{u^2 \sin 2\theta}{g} = \frac{u^2}{2g} \]

4. **Simplify the Equation**:
We can cancel \( u^2 \) and \( g \) (assuming \( u \neq 0 \) and \( g \neq 0 \)):
\[ \sin 2\theta = \frac{1}{2} \]

5. **Find the Angle**:
The sine function equals \( \frac{1}{2} \) at specific angles:
\[ 2\theta = 30^\circ \quad \text{or} \quad 2\theta = 150^\circ \]
Therefore, dividing by 2 gives:
\[ \theta = 15^\circ \quad \text{or} \quad \theta = 75^\circ \]

6. **Select the Correct Angle**:
The problem asks for the angle of projection. Both \( 15^\circ \) and \( 75^\circ \) are valid, but if we consider typical projectile motion scenarios, we often select the larger angle for maximum range. Thus, the angle of projection is:
\[ \theta = 75^\circ \]

### Final Answer:
The angle of projection is \( \theta = 75^\circ \).