Question

If two spheres are projected at angles \(45^{\circ }\) and \(60^{\circ }\) with horizontal and the maximum heights reached are same. The ratio of their initial velocities is (\(\sin 45^{\circ }=\frac{1}{\sqrt{2}},\,\sin 60^{\circ }=\frac{\sqrt{3}}{2}\))

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

Hint:

Use the maximum height formula for each projectile and equate the two heights.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The maximum height of a projectile is \(H=\dfrac{u^{2}\sin^{2}\theta}{2g}\).

Step 2: Equate the Heights
Since the heights are equal, \(u_1^{2}\sin^{2}45^{\circ}=u_2^{2}\sin^{2}60^{\circ}\).

Step 3: Substitute the Sines
\(u_1^{2}\cdot\dfrac{1}{2}=u_2^{2}\cdot\dfrac{3}{4}\).

Step 4: Find the Ratio
\(\dfrac{u_1^{2}}{u_2^{2}}=\dfrac{3}{2}\), so \(u_1:u_2=\sqrt{3}:\sqrt{2}\).

Final Answer:
The ratio of initial velocities is root 3 to root 2. \[ \boxed{\text{(D) }\sqrt{3}:\sqrt{2}} \]