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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Motion in a Plane

The path followed by a body projected along y axis is given by $y=\sqrt 3 x-\large\frac{1}{2}$$ x^2$. If $g=10 m/s^2$ Then initial velocity of projectile is

\[(a)\;3 \sqrt {10} \; m/s \quad (b)\;2 \sqrt {10} \;m/s \quad (c)\;10 \sqrt {3} \;m/s \quad(d)\;10 \sqrt {2} \; m/s \]

1 Answer

Comparing $y=\sqrt 3 x-\large\frac{1}{2} $$x^2$
with $y=\tan\theta  x-\large\frac{1}{2} \frac{g}{u^2 \cos ^2 \theta}$$x^2$
$\tan \theta=\sqrt 3=>\theta=60 ^{\circ}$
and $\large\frac{1}{2}=\frac{1}{2} \frac{g}{u^2 \cos ^2 \theta}$
=>$u^2\cos ^2 \theta=g$
$u^2\cos ^2 60=10$
$u^2.\large\frac{1}{4}=10$
$u^2=40$
$u=2 \sqrt {10}$
Hence b is the correct answer.

 

answered Jul 3, 2013 by meena.p 1 flag
edited Aug 14, 2014 by thagee.vedartham
 

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