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A cricket ball is thrown by a player at a speed of 20 m/s in a direction $30^\circ$ above the horizontal. The maximum height attained by the ball during its motion is ($g=10$ m/s$^2$)
Detailed Solution
$H=\frac{u^2\sin^2\theta}{2g}=\frac{(20)^2\times(1/2)^2}{2\times10}=5$ m.