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: Understanding the Concept
The maximum height depends on the vertical component of initial velocity: \(H=\dfrac{(u\sin\theta)^2}{2g}\).

Step 2: Key Formula or Approach
Equal heights mean equal vertical components: \(u_A\sin45^{\circ}=u_B\sin60^{\circ}\).

Step 3: Detailed Explanation
\[ \frac{u_A}{u_B}=\frac{\sin60^{\circ}}{\sin45^{\circ}}=\frac{\sqrt3/2}{1/\sqrt2}=\frac{\sqrt3\,\sqrt2}{2}=\frac{\sqrt3}{\sqrt2} \]
The ratio is \(\sqrt3:\sqrt2\).

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