Question

A particle is aimed at a mark which is in the same horizontal plane as that of the point of projection. It falls $10m$ short of the target when it is projected with an elevation of ${75}^{\circ }$ and falls $10m$ ahead of the target when it is projected with an elevation of ${45}^{\circ }$. If $\theta$ is the correct elevation so that it exactly hits the target, find the value of $\sin 2\theta$ (for the same velocity of projection)

• A. $0.5$
• B. $0.1$
• C. $0.75$
• D. $0.25$

Answer 1

The correct option is C $0.75$

Let $R$ be the range of the particle, $R=\frac{u^2\sin 2\theta }{g}$
For the condition given in question,
$R-10=\frac{u^2\sin {150}^{\circ }}{g}=\frac{u^2}{2g}....\left(1\right)R+10=\frac{u^2\sin {90}^{\circ }}{g}=\frac{u^2}{g}....\left(2\right)$
From $\left(1\right)$ & $\left(2\right)$
$(R-10)=(R+10)\frac{1}{2}$
$\Rightarrow 2R-20=R+10\Rightarrow R=30m$
i.e. $\frac{u^2}{g}=R+10=40m$
If particle is projected with correct elevtion,
$R=\frac{u^2}{g}\sin 2\theta =40\sin 2\theta =30\Rightarrow \sin 2\theta =\frac{3}{4}$