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Motion in a Plane
Time taken to reach A to B for a projectile is 't'. Then the distance AB is equal to
$ a)\;\large\frac{ut}{\sqrt 3} \\ b)\;\large\frac{\sqrt 3 ut}{2}\\ c)\;\sqrt ut \\ d)\;2 ut$
jeemain
physics
class11
unit2
kinematics
motion-in-a-plane
projectile-motion
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asked
Jun 26, 2013
by
meena.p
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Aug 11, 2014
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1 Answer
Horizontal component of u is $u \cos 60$
$=\large\frac{u}{2}$
$AC=\large\frac{u}{2}$$t=\large\frac{ut}{2}$
$AB=AC \sec 30$
$\qquad=\large\frac{ut}{2} \frac{2}{\sqrt 3}$
$\qquad=\large\frac{ut}{\sqrt 3}$
Hence a is the correct answer.
answered
Jun 26, 2013
by
meena.p
edited
Jan 25, 2014
by
meena.p
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