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A particle is projected with an angle of projection $\theta$ to the horizontal line passing through the points $( P , Q )$ and $( Q , P )$ referred to horizontal and vertical axes (can be treated as $x$-axis and $y$-axis respectively).
The angle of projection can be given by
$\tan ^{-1}\left[\frac{ P ^2+ PQ + Q ^2}{ PQ }\right]$
$\tan ^{-1}\left[\frac{ P ^2+ Q ^2- PQ }{ PQ }\right]$
$\tan ^{-1}\left[\frac{ P ^2+ Q ^2}{2 PQ }\right]$
$\sin ^{-1}\left[\frac{ P ^2+ Q ^2+ PQ }{2 PQ }\right]$
Solution
(a)
The general equation of path for projectile motion is
$y=x \tan \theta-\frac{g x^2}{2 u^2(\cos \theta)^2}$
Now since the above equation passes through (P,Q) and (Q,P) so we will get two equations as
$P=Q \tan \theta-\frac{g Q^2}{2 u^2(\cos \theta)^2}$
$Q=P \tan \theta-\frac{g P^2}{2 u^2(\cos \theta)^2}$
Solving equation $1$ and $2$ simultaneously we get
$\theta=\tan ^{-1} \frac{\left[ P ^2+ Q ^2+ PQ \right]}{[ PQ ]}$
