Question

The equation of motion of a projectile is $y=ax-b{x}^2$, where $a$ and $b$ are constants of motion. Match the quantities in Column I with the relations in Column II.

Column I		Column II
(A)	The initial velocity of projection	(p)	$\frac{a}{b}$
(B)	The horizontal range of projectile	(q)	$a\sqrt{\frac{2}{bg}}$
(C)	The maximum vertical height attained by projectile	(r)	$\frac{a^2}{4b}$
(D)	The time of flight of projectile	(s)	$\sqrt{\frac{g\left(1+a^2\right)}{2b}}$

• A. $A-p$, $B-q$, $C-r$, $D-s$
• B. $A-s$, $B-p$, $C-q$, $D-r$
• C. $A-s$, $B-p$, $C-r$, $D-q$
• D. $A-p$, $B-s$, $C-r$, $D-q$

Answer 1

Answer: (C)

Comparing given equation,
$y=ax-b{x}^2$ with
the equation of projectile motion $y=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }$,
we get, $tan\theta =a...(i)$
and $\frac{g}{2u^{2}co{s}^2\theta }=b$
$\therefore \frac{gse{c}^2\theta }{2u}=b$ or
$\frac{g\left(1+ta{n}^2\theta \right)}{2u^2}=b$
or $u^2=\frac{g\left(1+ta{n}^2\theta \right)}{2b}$
$=\frac{g\left(1+a^2\right)}{2b}$ (Using (i))
or $u=\sqrt{\left[\frac{g\left(1+a^2\right)}{2b}\right]}$; $A-s$
(B) Horizontal range,
$R=\frac{u^{2}si{n}^{2}2\theta }{g}$
$=\frac{2u^{2}sin\theta cos\theta }{g}$
$R=\frac{2u^{2}co{s}^2\theta }{g}\times tan\theta =\frac{a}{b}$ (Using (i) and (ii))
$B-p$
(C) Maximum height,
$H=\frac{u^{2}si{n}^{2}2\theta }{2g}$
$H=\frac{u^{2}co{s}^2\theta }{2g}\times ta{n}^2\theta =\frac{2u^{2}co{s}^2\theta }{4g}\times ta{n}^2\theta =\frac{a^2}{4b}$ (Using (i) and (ii))
$C-r$
(D) From eqn. (ii),
$u^{2}co{s}^2\theta =\frac{g}{2b}$ or $ucos\theta =\sqrt{\frac{g}{2b}}...\left(iii\right)$
Time of flight $=\frac{2usin\theta }{g}=\frac{2ucos\theta }{g}\times tan\theta$
=$\frac{2}{g}\sqrt{\frac{g}{2b}}\times a=a\sqrt{\frac{2}{bg}}$ (Using (i) and (iii)
$D-q$