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A particle is projected with velocity u so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as (nu²)/(25g), where value of n is (Given, 'g' is the acceleration due to gravity.)

Asked in JEE Main 3rd April 2nd Shift 2025 · Maximum height

Answer: (4) 24

Step-by-step solution

R = 3H ⇒ (u² sin 2θ)/g = (3u² sin²θ)/(2g) ⇒ 4 sin θ cos θ = 3 sin²θ ⇒ tan θ = 4/3.

sin 2θ = (2 tan θ)/(1 + tan²θ) = (8/3)/(25/9) = (24)/(25).

R = (24u²)/(25g), so n = 24.

Why the other options are wrong

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