Question

A particle is projected from the ground with an initial speed of u at an angle \( \theta \) with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is:

• A. \( u \cos \theta \)
• B. \( \frac{u}{2}\sqrt{1 + \cos^2 \theta} \)
• C. \( \frac{u}{2}\sqrt{1 + 2\cos^2 \theta} \)
• D. \( \frac{u}{2}\sqrt{1 + 3\cos^2 \theta} \)

Hint:

Use identity: \( \sin^2\theta = 1 - \cos^2\theta \) to simplify expressions.

Answer 1

The Correct Option is D

Concept: \[ \vec{v}_{avg} = \frac{\text{displacement}}{\text{time}} \]

Step 1: Time to reach highest point.
\[ t = \frac{u\sin\theta}{g} \]

Step 2: Displacement components.
Horizontal: \[ x = u\cos\theta \cdot t = \frac{u^2 \sin\theta \cos\theta}{g} \] Vertical: \[ y = \frac{(u\sin\theta)^2}{2g} = \frac{u^2 \sin^2\theta}{2g} \]

Step 3: Magnitude of displacement.
\[ R = \sqrt{x^2 + y^2} = \frac{u^2}{g}\sqrt{\sin^2\theta \cos^2\theta + \frac{1}{4}\sin^4\theta} \] \[ = \frac{u^2}{g} \cdot \frac{1}{2}\sqrt{4\sin^2\theta \cos^2\theta + \sin^4\theta} \] \[ = \frac{u^2 \sin\theta}{2g}\sqrt{4\cos^2\theta + \sin^2\theta} \]

Step 4: Average velocity.
\[ v_{avg} = \frac{R}{t} = \frac{\frac{u^2 \sin\theta}{2g}\sqrt{4\cos^2\theta + \sin^2\theta}}{\frac{u\sin\theta}{g}} \] \[ = \frac{u}{2}\sqrt{4\cos^2\theta + \sin^2\theta} \] \[ = \frac{u}{2}\sqrt{1 + 3\cos^2\theta} \]