The equation of motion of a projectile is y = ax$-{\mathrm{bx}}^2$ where a and b are constants of motion. Match the quantities of Column - I with the relations of Column - II and mark the correct choice from the given codes.

	Column-I		Column-II
(i)	The initial velocity of projection	(p)	$\frac{\mathrm{a}}{\mathrm{b}}$
(ii)	The horizontal range of projectile	(q)	a$\sqrt{\frac{2}{bg}}$
(iii)	The maximum vertical height attained by projectile	r)	$\frac{a^2}{4b}$
(iv)	The time of flight of projectile	(s)	$\sqrt{\frac{g(1+a^2)}{2b}}$

Codes

Comparing given equation, y = ax$-{\mathrm{bx}}^2$ with the equation of projectile motion

y = x tan $\mathrm{\theta }- \frac{{\mathrm{gx}}^2}{2{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}$

We get, $\tan \mathrm{\theta } = \mathrm{a}$-------(i)

       $\frac{\mathrm{g} {\sec }^2\mathrm{\theta }}{2{\mathrm{u}}^2} = \mathrm{b}$----(ii)

or ${\mathrm{u}}^2 = \frac{\mathrm{g}(1+{\tan }^2\mathrm{\theta })}{2\mathrm{b}} = \frac{\mathrm{g}(1+{\mathrm{a}}^2)}{2\mathrm{b}} (\mathrm{Using} (\mathrm{i})$

or u = $\sqrt{\left[\frac{g(1+a^2)}{2b}\right]}$

Here i-s.

Horizontal range = $\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}} = \frac{2{\mathrm{u}}^2\mathrm{sin\theta } \mathrm{cos\theta } }{\mathrm{g}}$

                          = $$\begin{gathered} = \frac{2{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}{\mathrm{g}}\times \mathrm{tan\theta } = \frac{\mathrm{a}}{\mathrm{b}} \\ (\mathrm{Using} (\mathrm{i}) \mathrm{and} (\mathrm{ii}\left)\right) \end{gathered}$$

Hence ii-p.

Maximum height = $\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}} = \frac{{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}{2\mathrm{g}}\times {\tan }^2\mathrm{\theta }$

                          = $$\begin{gathered} \frac{2{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}{4\mathrm{g}}\times {\tan }^2\mathrm{\theta } = \frac{{\mathrm{a}}^2}{4\mathrm{b}} \\ (\mathrm{Using} (\mathrm{i}) \mathrm{and} (\mathrm{ii}\left)\right) \end{gathered}$$

Hence iii-r.

From (ii), b = $\frac{\mathrm{g}}{2{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}$

or ${\mathrm{u}}^2{\cos }^2\mathrm{\theta } = \frac{\mathrm{g}}{2\mathrm{b}} \mathrm{or} \mathrm{u} \cos \mathrm{\theta } = \sqrt{\frac{\mathrm{g}}{2\mathrm{b}}} \left(\mathrm{iii}\right)$

Time of flight = $\frac{2\mathrm{u} \sin \mathrm{\theta }}{\mathrm{g}}=\frac{2}{g}u \cos \theta \times \mathrm{tan\theta }$

                    = $$\begin{gathered} \frac{2}{\mathrm{g}}\sqrt{\frac{\mathrm{g}}{2\mathrm{b}}}\times \mathrm{a} = \mathrm{a}\sqrt{\frac{2}{\mathrm{bg}}} \\ (\mathrm{Using} (\mathrm{i}) \mathrm{and} (\mathrm{iii}\left)\right) \end{gathered}$$

Hece iv-q