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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 the maximum height.

$\begin{array}{1 1}(A)\;100m \\(B)\;25 m \\(C)\;300m\\(D)\;2m \end{array} $

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1 Answer

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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 the angle of projection to be $ \theta =45$
Maximum height is given by
$H= \large\frac{v_0^2 \sin ^2 \theta }{g} $
$\quad= \large\frac{1000( 1/ \sqrt 2 )^2}{2(10)}$
$\qquad= 25 m$
Hence B is the correct answer.

 

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

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