Question

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