Question

Two spheres are projected at angles \(30^{\circ}\) and \(45^{\circ}\) with the horizontal. The maximum height reached by both is same. The ratio of their initial velocities is, \((sin45^{\circ} = \frac{1}{\sqrt{2}},sin30^{\circ} = 0.5)\)

• A. \(2:3\)
• B. \(\sqrt{2}:1\)
• C. \(3:1\)
• D. \(\sqrt{2}:\sqrt{3}\)

Hint:

Maximum height depends on the vertical part of the speed, u sin(theta).

Answer 1

The Correct Option is B

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

Step 2: Set equal
\[ u_1\sin30^\circ=u_2\sin45^\circ \]

Step 3: Solve for the ratio
\[ \frac{u_1}{u_2}=\frac{\sin45^\circ}{\sin30^\circ}=\frac{1/\sqrt2}{1/2}=\frac{2}{\sqrt2}=\sqrt2 \]
So \(u_1:u_2=\sqrt2:1\).

Step 4: Check the options
The ratio 2:3 and 3:1 do not come from the sine values. The ratio \(\sqrt2:\sqrt3\) would come from using \(\tan\) or mixing in \(\cos\). The correct one is option (B).

Final Answer:
Equal heights need equal vertical speeds, so u1 : u2 = sin45 : sin30 = root 2 : 1, option (B). \[ \boxed{\sqrt2:1} \]