Question

The equation of motion of a projectile is $y = ax - bx^{2}$, where a and b are constants of motion. Match the quantities of Column I with the relations of 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)	$$\sqrt{\frac{g(1 + a^{2})}{2b}}$$

• 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

Answer: (C)

A – s, B – p, C – r, D – q

Answer 2

Comparing given equations , $y = ax - bx^{2}$with the equations of projectile motions

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

We get,$\tan\theta = a$ …..(i)

And $\frac{g}{2u^{2}\cos^{2}\theta} = b$ …. (ii)

$\frac{g\sec^{2}\theta}{2u^{2}} = b$or $\frac{g(1 + \tan^{2}\theta)}{2u^{2}} = b$

Or $u^{2} = \frac{g(1 + \tan^{2}\theta)}{2b} = \frac{g(1 + a^{2})}{2b}$ (using (i))

Or $u = \sqrt{\left\lbrack \frac{g(1 + a^{2})}{2b} \right\rbrack}$A –s

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

$\frac{2u^{2}\cos^{2}\theta}{g} \times \tan\theta = \frac{a}{b}$ (Using (i) and (ii)) B – p

(3) Maximum height$= \frac{u^{2}\sin^{2}\theta}{2g}$

$= \frac{u^{2}\cos^{2}\theta}{2g} \times \tan^{2}\theta$

$\frac{2u^{2}\cos^{2}\theta}{4g} \times \tan^{2}\theta = \frac{a^{2}}{4b}$ (Using (i) and (ii))

C – r

(4) From (ii),

$b = \frac{g}{2u^{2}\cos^{2}\theta},$

Or $u^{2}\cos^{2}\theta = \frac{g}{2b}oru\cos\theta = \sqrt{\frac{g}{2b}}$ …… (iii)

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