Single Choice

A particle is projected with an angle of projection 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$$\tan^{-1}\left[\dfrac{P^2+PQ+Q^2}{PQ}\right]$$
Correct Answer
B$$\tan^{-1}\left[\dfrac{P^2+Q^2-PQ}{PQ}\right]$$
C$$\tan^{-1}\left[\dfrac{P^2+Q^2}{2PQ}\right]$$
D$$\sin^{-1}\left[\dfrac{P^2+Q^2+PQ}{2PQ}\right]$$

Solution

Given that,
A particle is projected with an angle of projection to the horizontal line passing through the points $$(P,Q)$$ and $$(Q,P)$$.

The general equation of path for projectile motion is
$$y = x\ tan\theta+\dfrac{gx^2}{2u^2(\cos\theta)^2}$$

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

$$P=Q\tan\theta-\dfrac{gQ^2}{2u^2(\cos\theta)^2}$$......(I)
$$Q=P\tan\theta-\dfrac{gP^2}{2u^2(\cos\theta)^2}$$.....(II)

From equation (I) and (II) after solving
$$\theta =\tan^{-1}[ \dfrac{P^2+PQ+Q^2}{PQ}]$$
So, The angle of projection is $$\theta =\tan^{-1}[ \dfrac{P^2+PQ+Q^2}{PQ}]$$
Hence, A is correct option.

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