Question

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

• A. ${tan}^{-1}\left[\begin{array}{c}\frac{P^2+PQ+Q^2}{PQ}\end{array}\right]$
• B. ${tan}^{-1}\left[\begin{array}{c}\frac{P^2+Q^2-PQ}{PQ}\end{array}\right]$
• C. ${tan}^{-1}\left[\begin{array}{c}\frac{P^2+Q^2}{2PQ}\end{array}\right]$
• D. ${tan}^{-1}\left[\begin{array}{c}\frac{P^2+Q^2+PQ}{2PQ}\end{array}\right]$

Answer 1

The correct option is A ${tan}^{-1}\left[\begin{array}{c}\frac{P^2+PQ+Q^2}{PQ}\end{array}\right]$

The general equation of path for projectile motion is

$y=x\tan \theta -\frac{g{x}^2}{2u^2(\cos \theta {)}^2}$

Now since the above equation passes through (P,Q) and (Q,P) so we will get two equations as

$P=Q\tan \theta -\frac{g{Q}^2}{2u^2(\cos \theta {)}^2}$ ..............(1)

$Q=P\tan \theta -\frac{g{P}^2}{2u^2(\cos \theta {)}^2}$.....................(2)

Solving equation 1 and 2 simultaneously we get

$\theta ={\tan }^{-1}\frac{[P^2+Q^2+PQ]}{\left[PQ\right]}$