Question

For angles of projection of a projectile at angles (45°θ) and (45° +-θ), the horizontal ranges described by the projectile are in the ratio of :

• A. 1 : 1
• B. 2 : 3
• C. 1 : 2
• D. 3 : 2

Answer 1

The Correct Option is A

To determine the ratio of horizontal ranges for a projectile launched at angles 45^\circ - \theta and 45^\circ + \theta, we follow these steps:

Step 1: Understanding the Projectile Motion Formula

The horizontal range R of a projectile is given by the formula:

R = \frac{u^2 \sin 2\alpha}{g}

where:

• u is the initial velocity.
• \alpha is the angle of projection.
• g is the acceleration due to gravity.

Step 2: Calculate the Ranges for Given Angles

Let's denote:

• \alpha = 45^\circ - \theta
• \beta = 45^\circ + \theta

The range for angle \alpha is:

R_1 = \frac{u^2 \sin 2(45^\circ - \theta)}{g}

The range for angle \beta is:

R_2 = \frac{u^2 \sin 2(45^\circ + \theta)}{g}

Step 3: Use the Trigonometric Identity

We know:

\sin(90^\circ - x) = \cos(x)

Thus:

\sin 2(45^\circ - \theta) = \sin(90^\circ - 2\theta) = \cos(2\theta)

And:

\sin 2(45^\circ + \theta) = \sin(90^\circ + 2\theta) = \cos(-2\theta) = \cos(2\theta)

Step 4: Conclusion and Answer

Since both expressions become \cos(2\theta), we find that:

R_1 = R_2

Therefore, the ratio of the ranges is:

R_1 : R_2 = 1 : 1

Final Answer: The correct option is 1 : 1.