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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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The equation of trajectory of a projectile is given by $y=x -\large\frac{x^2}{100}$ Find velocity of projection .

$\begin{array}{1 1}(A)\;\frac{1}{100} \\(B)\;10 \sqrt {10} ms^{-1} \\(C)\;\frac{1}{300}\\(D)\;2 \end{array} $

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Comparing $y=x -\large\frac{x^2}{100}$
with $y=\tan\theta  x-\large\frac{1}{2} \frac{g}{u^2 \cos ^2 \theta}$$x^2$
we get that the value of $\theta=45$
the velecity of projection is given by
$v_0^2 =\large\frac{100 g}{2 \cos ^2 \theta_0} =\frac{100(10)}{2(1/ \sqrt 2 )^2}$$=1000(m/s^2)$
$\qquad= v_0 = 10 \sqrt {10} ms^{-1}$
Hence B is the correct answer.


answered Jul 15, 2014 by meena.p
edited Aug 14, 2014 by thagee.vedartham

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