Question

A body is thrown up with a speed \( u \), at an angle of projection \( \theta \). If the speed of the projectile becomes \( \frac{u}{\sqrt{2}} \) on reaching the maximum height, the maximum vertical height attained by the projectile is

• A. $\frac{u^2}{4g}$
• B. $\frac{u^2}{3g}$
• C. $\frac{u^2}{2g}$
• D. $\frac{u^2}{g}$
• E. $\frac{2u^2}{g}$

Hint:

At highest point, only horizontal velocity remains.

Answer 1

The Correct Option is A

Concept: Velocity components in projectile motion
At maximum height:
• Vertical velocity = 0
• Only horizontal component remains ---

Step 1: Initial velocity components
\[ u_x = u \cos\theta, \quad u_y = u \sin\theta \] ---

Step 2: Velocity at highest point
\[ v = u \cos\theta \] Given: \[ u \cos\theta = \frac{u}{\sqrt{2}} \] \[ \cos\theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = 45^\circ \] ---

Step 3: Maximum height formula
\[ H = \frac{u^2 \sin^2\theta}{2g} \] Substitute: \[ \sin^2 45^\circ = \frac{1}{2} \] \[ H = \frac{u^2}{2g} \cdot \frac{1}{2} = \frac{u^2}{4g} \] --- Final Answer: \[ \boxed{\frac{u^2}{4g}} \]