Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A body is projected at such an angle that the horizontal range is three times the greatest height. The angle of projection is:

25°8′

33°7′

42°8′

53°8′

Formulas: Maximum height ( $H$ ) is given by $H = \frac{u^2 \sin^2 \theta}{2g}$ . Horizontal range ( $R$ ) is given by $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$ .
Given Condition: $R = 3H$ .
Substitution: $\frac{2u^2 \sin \theta \cos \theta}{g} = 3 \times \frac{u^2 \sin^2 \theta}{2g}$
Simplification: Cancel $u^2$ , $g$ , and one $\sin \theta$ (since $\theta \neq 0$ ): $2 \cos \theta = \frac{3}{2} \sin \theta$ $\frac{\sin \theta}{\cos \theta} = \frac{4}{3}$ $\tan \theta = \frac{4}{3} = 1.333...$
Calculate Angle: $\theta = \tan^{-1}(1.333) \approx 53.13^{\circ}$ . Converting decimals to minutes: $0.13^{\circ} \approx 8'$ . Therefore, $\theta = 53^{\circ}8'$ .

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