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)a/b
(B)The horizontal range of projectile	(q)a$\frac{2}{bg}$
(C)The maximum vertical height attained by projectile	(r)$a^2$/4b
(D)The time of flight of projectile	(s)$\frac{g(1+a^2)}{2b}$

• 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

The correct option is C A-s,B-p,C-r,D-q

Comparing given equation, y = ax - bx${}^2$ with

the equation of projectile motion y = x tan $\theta$ - $\frac{g{x}^2}{2u^{2}co{s}^2\theta }$,

we get, tan$\theta$ = a---(i) and $\frac{g}{2u^{2}co{s}^2\theta }$ = b----(ii)

$\therefore$ $\frac{gse{c}^2\theta }{2u^2}$= b or $\frac{g(1+ta{n}^2\theta )}{2u^2}$ =b

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

$A-s$

(B) Horizontal range, R =$\frac{u^{2}si{n}^2\theta }{g}$=$\frac{u^{2}sin\theta cos\theta }{g}$

R= $\frac{u^{2}co{s}^2\theta }{g}$X tan$\theta$= a/b (Using (i) and (ii))

$B-p$

(C) Maximum height, H = $\frac{u^{2}si{n}^2\theta }{g}$

H= $\frac{u^{2}co{s}^2\theta }{2g}$X tan${}^2\theta$= $\frac{u^{2}co{s}^2\theta }{4g}$ X tan${}^2\theta$= $\frac{a^2}{4b}$

$C-r$ (Using (i) and (ii))

(D) From eqn. (ii),

$u^{2}co{s}^2\theta$= g/2b or ucos$\theta$= $\sqrt{\frac{g}{2b}}$-----(iii)

Time of flight= $\frac{2usin\theta }{g}$ = $\frac{2ucos\theta }{g}$ X tan$\theta$

= $\frac{2}{g}\sqrt{\frac{g}{2b}}Xa$= $a\sqrt{\frac{2}{bg}}$ (Using (i) and (iii))

$D-q$