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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.

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

Can you answer this question?

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
edited Aug 14, 2014