A particle is projected with velocity u so that its horizontal ra… - Vidyadip

MCQ

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).

A
6
B
18
C
12
D
24

✓

Answer

D. 24
Range $=3 \mathrm{H}_{\text {max }}$
$\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{~g}}=\frac{3 \mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}$
$2 \sin \theta \cos \theta=\frac{3}{2} \sin ^{2} \theta$
$\tan \theta=\frac{4}{3} \Rightarrow \theta=53^{\circ}$
$R=\frac{u^{2}\left(2 \times \frac{3}{5} \times \frac{4}{5}\right)}{g} \Rightarrow \frac{24 u^{2}}{25 g}$

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