Question

Two parabolic paths of angles $\alpha$ and $\beta$ of projection aimed at a target on the horizontal plane through, $O$, fall $p$ units short and the other $p$ units far from the target. If $\theta$ is the correct angle of projection so as to hit the target, then $\sin 2\theta$ is equal to

• A. $\frac{1}{2}(\tan \alpha +\tan \beta )$
• B. $\frac{1}{2}(\tan \alpha -\tan \beta )$
• C. $\frac{1}{2}(\sin 2\alpha +\sin 2\beta )$
• D. $\frac{1}{2}(\sin 2\alpha -\sin 2\beta )$

Answer 1

The correct option is C $\frac{1}{2}(\sin 2\alpha +\sin 2\beta )$

From the figure
$OA=R-p=\frac{u^2\sin 2\alpha }{g}...\left(i\right)$
$OC=R+p=\frac{u^2\sin 2\beta }{g}...\left(ii\right)$
$OB=R=\frac{u^2\sin 2\theta }{g}...\left(iii\right)$
adding (i) and (ii)
$\Rightarrow 2R=\frac{u^2}{g}[\sin 2\alpha +\sin 2\beta ]...\left(iv\right)$
$\therefore \frac{2R}{R}=\frac{\sin 2\alpha +\sin 2\beta }{\sin 2\theta }$
$\Rightarrow \sin 2\theta =\frac{1}{2}(\sin 2\alpha +\sin 2\beta )$