A particle is projected on an inclined plane which is inclined at 300 with the horizontal as shown in figure. Initial speed of the particle is v0 , and inclined plane is sufficiently large. Match the Column- I and Column – II. Column-IColumn-IIa)Range on the inclined planep)v0g3b)Velocity of the particle is parallel to the inclined plane at timeq)2v0g3c)Time after which particle strikes the plane isr)4v023gd)For the given velocity maximum range on the inclined plane(angle of projection changing) s)2v023g

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A particle is projected on an inclined plane which is inclined at $30^{0}$ with the horizontal as shown in figure. Initial speed of the particle is $v_{0}$ , and inclined plane is sufficiently large. Match the Column- I and Column – II.

Column-I		Column-II
a)	Range on the inclined plane	p)	$\frac{v_{0}}{g \sqrt{3}}$
b)	Velocity of the particle is parallel to the inclined plane at time	q)	$\frac{2 v_{0}}{g \sqrt{3}}$
c)	Time after which particle strikes the plane is	r)	$\frac{4 v_{0}^{2}}{3 g}$
d)	For the given velocity maximum range on the inclined plane (angle of projection changing)	s)	$\frac{2 {v_{0}}^{2}}{3 g}$

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a-s; b-p; c-q; d-s;

a-s; b-q; c-q; d-s;

a-s; b-p; c-p; d-s;

a-r; b-p; c-q; d-s;

answer is D.

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Detailed Solution

Range $R = \frac{2 V_{0}^{2} \sin 30^{0} \cos 60^{0}}{g \cos^{2} 30^{0}}$
Time when velocity is parallel to plane $V_{_{0}} \sin 30^{0} - g \cos 30^{0} t = 0$
$t = \frac{V_{0}}{g \sqrt{3}}$
Time of flight $T = \frac{2 V_{0} \sin α}{g \cos β}$
Maximum range $R = \frac{v_{0}^{2}}{g (1 + \sin β)}$