Question

A particle is aimed at a mark which is in the same horizontal plane as that of point of projection. If it falls $10m$ short of the target when it is projected with in elevation of ${75}^o$ and fall $10m$ ahead of the target when it is projected with an elevation of ${45}^o$. 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. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A $\frac{3}{4}$

Case I: Range, $R-10=\frac{u^2\sin {150}^o}{g}$..........(1)
Case II: Range,
$R+10=\frac{u^2\sin 90}{g}=\frac{u^2}{g}$ ..........(2)
Case III: Range,
$R=\frac{u^2\sin 2\theta }{g}$ ................(3)

From (1) & (2)
$(R-10)=(R+10)\frac{1}{2}$
$2R-20=R+10$
$R=30$
In equation (3), put $R=30$ and $\frac{u^2}{g}=40$, we get
$30=40\sin 2\theta$
$\Rightarrow \sin 2\theta =\frac{3}{4}$