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
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
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Same maximum height implies:
u² sin²θ = constant
Always check physical validity of the obtained angle.
Updated On: Mar 19, 2026
- 0^∘
- 60^∘
- 30^∘
- 45^∘
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The Correct Option is A
Solution and Explanation
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^∘
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