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Q. A cricket ball is thrown at a speed of $30\,ms^{-1}$ in a direction $30^°$ above the horizontal, the maximum height attained by the ball is

Motion in a Plane

Solution:

The maximum height is given by
$H=\frac{u^{2}\,sin^{2}\,\theta}{2g}$
$=\frac{\left(30\right)^{2}\,sin^{2}\,30^{°}}{2\times10}$
$=11.25\,m$