Question

Two bodies are projected from the same point with the same initial velocity 'u' making angles '$\theta$' and ($90^\circ-\theta$) with the horizontal in opposite directions. The horizontal distance between their positions when the bodies are at their maximum heights is

• A. $\frac{u^2}{2g}(\sin^2 \theta - \cos^2 \theta)$
• B. $\frac{u^2 \sin 2\theta}{2g}$
• C. $\frac{u^2}{g}$
• D. $\frac{u^2 \sin 2(90^\circ-\theta)}{g}$

Hint:

For complementary angles of projection ($\theta$ and $90-\theta$), the horizontal ranges are equal. The maximum heights occur at $R/2$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The two bodies are projected with complementary angles $\theta$ and $90^\circ-\theta$. For complementary angles, the horizontal range $R$ is the same. $R = \frac{u^2 \sin 2\theta}{g}$.
Step 2: Position of Maximum Height:
The horizontal distance to the maximum height for a projectile is half the range ($R/2$). Since they are projected in opposite directions from the same point: - Body 1 travels horizontal distance $x_1 = R/2$ to the right. - Body 2 travels horizontal distance $x_2 = R/2$ to the left.
Step 3: Calculate Separation:
The total horizontal distance between them when both are at their respective maximum heights is: \[ D = x_1 + x_2 = \frac{R}{2} + \frac{R}{2} = R \] \[ D = \frac{u^2 \sin 2\theta}{g} \]
Step 4: Check Options:
Option (D) is $\frac{u^2 \sin 2(90^\circ-\theta)}{g}$. Using $\sin(180^\circ - A) = \sin A$: \[ \sin 2(90^\circ-\theta) = \sin(180^\circ - 2\theta) = \sin 2\theta \] So, Option (D) simplifies to $\frac{u^2 \sin 2\theta}{g}$, which is the correct distance.