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: Understanding the Concept
With no air resistance the horizontal velocity of a projectile never changes, only the vertical component does.

Step 2: Initial horizontal component
The ball is hit at \(60^{\circ}\) with the vertical, so at \(30^{\circ}\) with the horizontal. So:
\[ v_x = v\cos 30^{\circ} = \frac{\sqrt3}{2}v \]

Step 3: Later instant
Now the velocity makes \(60^{\circ}\) with the horizontal, so \(v_x = u\cos 60^{\circ} = \dfrac{u}{2}\), where u is the speed.
\[ \frac{u}{2} = \frac{\sqrt3}{2}v \Rightarrow u = \sqrt3\,v \]
The speed is larger than v because the ball is lower than its launch height at that instant, so it has gained kinetic energy.

Final Answer:
The speed at that instant is \(\sqrt3 v\), option (B). \[ \boxed{\sqrt{3}\,v} \]