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 , where value of n is : (Given 'g' is the acceleration due to gravity). [20

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Solution

Q.
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 $\frac{n u^{2}}{25 g}$ , where value of n is : (Given 'g' is the acceleration due to gravity). [2025]

1 6

2 18

3 12

4 24

Ans.
(4)

Range R = $3 H_{\max}$

$\frac{u^{2} \sin 2 θ}{g} = \frac{3 u^{2} \sin^{2} θ}{2 g}$

$2 \sin θ \cos θ = \frac{3}{2} \sin^{2} θ$

$\tan θ = \frac{4}{3} \Rightarrow θ = 53 °$

$R = \frac{u^{2} (2 \times \frac{3}{5} \times \frac{4}{5})}{g} = \frac{24 u^{2}}{25 g}$

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