The magnitude of velocity of projectile and time are related as $\frac{v^2}{a^2}-{(t-\frac{b}{a})}^2=\frac{c^2}{a^2}$then  

Column-I	Column-II
a)Range	P) c
b)Height	Q) 2b/a
c)Time of flight	R) 2bc/a
d) Velocity at maximum height	S) b2/2a
	T) ab/c

$\frac{V^2}{a^2}-{\left(t-\frac{b}{a}\right)}^2=\frac{c^2}{a^2}$Standard equation is

$\begin{array}{r}{V}^2=u^2{\mathrm{Cos}}^2\theta +(u\sin \theta -gt{)}^2----\left(1\right)\\ \end{array}$

$V^2=c^2+{\left[a\left(t-\frac{b}{a}\right)\right]}^2$

$V^2=c^2+(at-b{)}^2$----------------(2)

comparing  1 and 2

$\theta \cos u=c,\theta \sin u=b,g=a$

$Range =\frac{2u\sin \theta }{g}u\cos \theta =\frac{2bc}{a}$

$\mathrm{Maximum} \mathrm{Height}=\mathrm{H}=\frac{{\left(\mathrm{usin\theta }\right)}^2}{2\mathrm{g}}=\frac{{\mathrm{b}}^2}{2\mathrm{a}}$

$\mathrm{Time} \mathrm{of} \mathrm{flight}=\frac{2\mathrm{usin\theta }}{\mathrm{g}}=\frac{2\mathrm{b}}{\mathrm{g}}$