Question

Two balls A and B are projected at an angle of \(45^{\circ}\) and \(60^{\circ}\) respectively, so that the maximum heights reached are same for both. The ratio of initial velocity of projection of ball A to that for ball B is
\((sin30^{\circ} = cos60^{\circ} = \frac{1}{2}, sin45^{\circ} = cos45^{\circ} = \frac{1}{\sqrt{2}}, sin60^{\circ} = cos30^{\circ} = \frac{\sqrt{3}}{2})\)

• A. \(2:\sqrt{3}\)
• B. \(\sqrt{3}:2\)
• C. \(\sqrt{2}:\sqrt{3}\)
• D. \(\sqrt{3}:\sqrt{2}\)

Hint:

Maximum height is \(H=\dfrac{u^2\sin^2\theta}{2g}\); equate for both balls.

Answer 1

The Correct Option is D

Step 1: Set the heights equal
$\dfrac{u_A^2\sin^245^{\circ}}{2g}=\dfrac{u_B^2\sin^260^{\circ}}{2g}$, so $\dfrac{u_A^2}{2}=\dfrac{3u_B^2}{4}$.

Step 2: Solve
$\dfrac{u_A^2}{u_B^2}=\dfrac32$, so $\dfrac{u_A}{u_B}=\dfrac{\sqrt3}{\sqrt2}$, option (D).

Final Answer:
The ratio is $\sqrt3:\sqrt2$, option (D). \[ \boxed{\sqrt3:\sqrt2} \]