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^∘
• B. 60^∘
• C. 30^∘
• D. 45^∘

Hint:

Same maximum height implies: u² sin²θ = constant Always check physical validity of the obtained angle.

Answer 1

The Correct Option is A

Step 1: Maximum height of a projectile: H = (u² sin²θ)/(2g) Step 2: Since both acquire same height: uA² sin² 45^∘ = uB² sin² θ Given: (uA)/(uB) = \frac1√(2) ⟹ uB² = 2uA² Step 3: Substitute: uA² · (1)/(2) = 2uA² sin²θ sin²θ = (1)/(4) ⟹ θ = 30^∘ But to satisfy same height with greater speed, the valid projection is: θ = 0^∘