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: Conservation of horizontal velocity:
Let $u_x$ be the horizontal component. It stays the same, so $u_x = v\sin60^\circ$, where 60 degrees is measured from the vertical.

Step 2: Equate:
$v\sin60^\circ = v'\cos60^\circ$, so $v' = v\tan60^\circ = \sqrt3\,v$ (B).

Final Answer:
$\sqrt3\,v$. \[ \boxed{\sqrt{3}\,v} \]