Question

A batsman hits a ball with a velocity 'v', making an angle of \(60^{\circ}\) with the vertical. After some time direction of velocity is making an angle of \(60^{\circ}\) with the horizontal. The speed of the ball at this instant is
\([cos(60^{\circ}) = \frac{1}{2},cos(30^{\circ}) = \frac{\sqrt{3}}{2}]\)

• A. \(\frac{\sqrt{3}}{2}v\)
• B. \(\sqrt{3}v\)
• C. \(\frac{v}{2}\)
• D. \(V\)

Hint:

The horizontal component of velocity stays constant in projectile motion.

Answer 1

The Correct Option is B

Step 1: Constant component
Only gravity acts, and it is vertical, so $v\cos\alpha$ is the same throughout.

Step 2: Equate
$v\cos30^{\circ} = u\cos60^{\circ}$.

Step 3: Solve
$u = v\cdot\dfrac{\sqrt3/2}{1/2} = \sqrt3\,v$. Option (B).

Final Answer:
Option (B). \[ \boxed{\sqrt{3}\,v} \]