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) for an oblique projectile, given that the horizontal range (R) is equal to the distance (d) through which a projectile falls freely from rest to acquire a velocity equal to the velocity of projection (u) in magnitude.

### Step 1: Understand the relationship between range and distance fallen
The horizontal range (R) of a projectile launched at an angle θ with an initial velocity u is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where g is the acceleration due to gravity.

### Step 2: Calculate the distance fallen to acquire velocity u
For a freely falling object from rest, the final velocity (v) after falling a distance (d) is given by the equation:
\[ v^2 = u^2 + 2gd \]
Since the initial velocity (u) is 0, we have:
\[ v^2 = 2gd \]
If we want the final velocity (v) to equal the projection velocity (u), we can substitute v with u:
\[ u^2 = 2gd \]
From this, we can express d as:
\[ d = \frac{u^2}{2g} \]

### Step 3: Set the two equations equal
According to the problem, the horizontal range (R) is equal to the distance fallen (d):
\[ \frac{u^2 \sin(2\theta)}{g} = \frac{u^2}{2g} \]

### Step 4: Simplify the equation
We can cancel \( u^2 \) and \( g \) from both sides:
\[ \sin(2\theta) = \frac{1}{2} \]

### Step 5: Solve for θ
The equation \( \sin(2\theta) = \frac{1}{2} \) has two solutions within the range of angles:
1. \( 2\theta = 30^\circ \) → \( \theta = 15^\circ \)
2. \( 2\theta = 150^\circ \) → \( \theta = 75^\circ \)

### Step 6: Determine the correct angle
Both angles (15° and 75°) are valid solutions. However, in the context of projectile motion, the angle of projection that is commonly used is 75° since it is closer to the maximum range angle (45°) and is more practical for achieving a longer range.

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

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