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JEE Main 2025 (3 Apr, Shift 2) · Time of Flight, Maximum Height & Range

Time of Flight, Maximum Height & Range question from JEE Main 2025 (3 Apr, Shift 2)

Q8JEE Main 2025 · 3 APR · SHIFT 2Medium

Time of Flight, Maximum Height & Range

A particle is projected with velocity uu so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as nu225g\frac{nu^2}{25g}, where value of nn is: (Given, ' gg ' is the acceleration due to gravity.)

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Answer: (d)

  1. RH=4tan⁡θ=3⇒tan⁡θ=43\frac{R}{H} = \frac{4}{\tan\theta} = 3 \Rightarrow \tan\theta = \frac{4}{3}
  2. So sin⁡θ=45\sin\theta = \frac{4}{5}, cos⁡θ=35\cos\theta = \frac{3}{5}, sin⁡2θ=2⋅45⋅35=2425\sin 2\theta = 2\cdot\frac{4}{5}\cdot\frac{3}{5} = \frac{24}{25}
  3. R=u2sin⁡2θg=24u225gR = \frac{u^2\sin 2\theta}{g} = \frac{24u^2}{25g}
  4. So n=24n = 24, answer (d).
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