Question

A projectile aimed at a mark, which is in the horizontal plane through the point of projection, falls "a" cm short of it when the elevation is $\alpha$ and goes "b" cm far when the elevation is $\beta .$ Show that, if the speed of projection is same in all the cases the proper elevation is:

• A. $\frac{1}{2}{\sin }^{-1}\left[\frac{b\sin 2\alpha +a\sin 2\beta }{a+b}\right]$
• B. $\frac{3}{2}{\cos }^{-1}\left[\frac{b\sin 2\alpha +a\sin 2\beta }{a+b}\right]$
• C. $\frac{1}{2}{\sin }^{-1}\left[\frac{a\sin 2\alpha +b\sin 2\beta }{a+b}\right]$
• D. $\frac{3}{2}{\cos }^{-1}\left[\frac{a\sin 2\alpha +b\sin 2\beta }{a+b}\right]$

Answer 1

The correct option is A $\frac{1}{2}{\sin }^{-1}\left[\frac{b\sin 2\alpha +a\sin 2\beta }{a+b}\right]$

$R=\frac{u^{2}sin2\theta }{g},R-a=\frac{u^{2}sin2\alpha }{g},R+b=\frac{u^{2}sin2\beta }{g}\frac{R-a}{R}=\frac{sin2\alpha }{sin2\theta },\frac{R+b}{R}=\frac{sin2\beta }{sin2\theta }\Rightarrow R=\frac{asin2\theta }{sin2\theta -sin2\alpha }=\frac{bsin2\theta }{sin2\beta -sin2\theta }\Rightarrow \frac{sin2\theta -sin2\alpha }{a}=\frac{sin2\beta -sin2\theta }{b}\Rightarrow (a+b)sin2\theta =asin2\beta +bsin2\alpha \Rightarrow \theta =\frac{1}{2}{sin}^{-1}\left(\frac{asin2\beta +bsin2\alpha }{a+b}\right)$