A particle is projected from level ground. Assuming projection point as origin, x-axis along horizontal and y-axis along vertically upwards. If particle moves in x-y plane and its path is given by $y = a x - b x^{2}$ where $a, b$ are positive constants. Then match the physical quantities given in column-I with the values given in column-II. (g in column II is acceleration due to gravity.)
Column – I Column – II
A) Horizontal component of velocity P) $\frac{a}{b}$
B) Time of flight Q) $\frac{a^{2}}{4 b}$
C) Maximum height R) $\sqrt{\frac{g}{2 b}}$
D) Horizontal range S) $\sqrt{\frac{2 a^{2}}{b g}}$

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A – P, B – Q, C – R, D – S

A – Q, B – R, C – P, D – S

A – R, B – P, C – S, D – Q

A – R, B – S, C – Q, D – P

answer is D.

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Detailed Solution

Equation of path is given as $y = a x - b x^{2}$
Comparing it with standard equation of projectile;
$y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ}$
$\tan θ = a, \frac{g}{2 u^{2} \cos^{2} θ} = b$
Horizontal component of velocity $= u \cos θ = \sqrt{\frac{g}{2 b}}$
$Time of flight T = \frac{2 u \sin θ}{g} = \frac{2 (u \cos θ) \tan θ}{g} = \frac{2 (\sqrt{\frac{g}{2 b}}) a}{g} = \sqrt{\frac{2 a^{2}}{b g}} Maximum height H = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{{[u \cos θ . \tan θ]}^{2}}{2 g} = \frac{{[\sqrt{\frac{g}{2 b}} . a]}^{2}}{2 g} = \frac{a^{2}}{4 b} Horizontal range R = \frac{u^{2} \sin 2 θ}{g} = \frac{2 (u \sin θ) (u \cos θ)}{g} = \frac{2 [\sqrt{\frac{g}{2 b}} . a] [\sqrt{\frac{g}{2 b}}]}{g} = \frac{a}{b}$

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Column – I		Column – II
A)	Horizontal component of velocity	P)	$\frac{a}{b}$
B)	Time of flight	Q)	$\frac{a^{2}}{4 b}$
C)	Maximum height	R)	$\sqrt{\frac{g}{2 b}}$
D)	Horizontal range	S)	$\sqrt{\frac{2 a^{2}}{b g}}$