The path of projectile is represented by  $y=Px-Q{x}^2$

	COLUMN-I		COLUMN-II
I	Range	P	P/Q
II	Maximum height	Q	P
III	Time of flight	R	$P^2/4Q$
IV	Tangent of angle of projection is	S	$\sqrt{\frac{2}{Qg}}P$
		T	Q

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