The path of projectile is represented by $y = P x - Q x^{2}$
COLUMN-I COLUMN-II
I Range P P/Q
II Maximum height Q P
III Time of flight R $P^{2} / 4 Q$
IV Tangent of angle of projection is S $\sqrt{\frac{2}{Q g}} P$
T Q

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$I - R, I I - P, I I I - Q, I V - S$

$I - P, I I - Q, I I I - R, I V - S$

$I - P, I I - R, I I I - S, I V - Q$

$I - Q, I I - P, I I I - R, I V - S$

answer is C.

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Detailed Solution

I $\to$ P, II $\to$ R, III $\to$ S, IV $\to$ Q
$y = P x - Q x^{2}$
Equation of trajectory of projectile motion is given by $y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ}$
On comparing, $\tan θ = p$ . So IV $\to$ Q
$\frac{g}{2 u^{2} \cos^{2} θ} = Q; u \cos θ = \sqrt{\frac{g}{2 Q}}$
Range, $R = \frac{2 u^{2}}{g} \sin θ \cos θ$
$∴ R = \frac{2}{g} \frac{g}{2 Q} P = \frac{P}{Q}$
$I \to P$ maximum height , $H = \frac{u^{2}}{2 g} \sin^{2} θ$
$∴ H = \frac{1}{2 g} \frac{g}{2 Q} \frac{\sin^{2} Q}{\cos^{2} θ} H = \frac{1}{2 Q} X \frac{P^{2}}{2 Q} = \frac{P^{2}}{4 Q^{2}}$
$I I \to R$ Time of light
$T = \frac{2 u \sin θ}{g} = \frac{2}{g} \sqrt{\frac{g}{2 Q}} p$
$T = \sqrt{\frac{2}{g Q}} P$