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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 – r, B – q, C – s, D – p

A – s, B – r, C – q, D – p

A – r, B – s, C – q, D – p

A – p, B – s, C – q, D – r

answer is C.

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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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