Question

A ball P is projected at an angle of \(60^{\circ}\) with the vertical with certain initial speed. Another ball Q of the same mass as that of ball P is projected vertically upwards with the same initial speed as that of P. At the highest point, the ratio of potential energy of ball P to that of ball Q is
\((sin30^{\circ} = 0.5)\)

• A. \(1:4\)
• B. \(4:1\)
• C. \(2:3\)
• D. \(3:2\)

Hint:

The horizontal component of velocity stays constant in projectile motion.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In projectile motion with no air resistance, the horizontal component of velocity does not change. Only the vertical component changes.

Step 2: Initial velocity:
The ball is hit at \(60^\circ\) to the vertical, so it makes \(30^\circ\) with the horizontal. Horizontal component \(= v\cos30^\circ = \frac{\sqrt3}{2}v\).

Step 3: Later instant:
Now the velocity makes \(60^\circ\) with the horizontal. If the speed is \(v'\), the horizontal component is \(v'\cos60^\circ = \frac{v'}{2}\).
\[ \frac{v'}{2} = \frac{\sqrt3}{2}v \Rightarrow v' = \sqrt3\,v \]
Check: the vertical component at this instant is \(\sqrt3 v\sin60^\circ = \frac32v\), larger than the initial \(\frac v2\), so the ball is at a lower height than the start on its way down.

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