Question

Two projectiles, A and B, are thrown with speeds in the ratio 1:√(2) and acquire the same heights. If A is thrown at an angle of 45° with the horizontal, the angle of projection of B will be

• A. \(0^\circ\)
• B. \(60^\circ\)
• C. \(30^\circ\)
• D. 45^∘

Hint:

Equal heights depend only on vertical component of velocity.

Answer 1

The Correct Option is C

Step 1: Maximum height:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Step 2: Since heights are equal:
\[ u_1^2 \sin^2 45^\circ = u_2^2 \sin^2 \theta \]
Step 3: Given \(u_1 : u_2 = 1 : \sqrt{2}\)
Step 4: Solving gives:
\[ \theta = 30^\circ \]