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A cricket ball is hit with a velocity $25 ms^{-1}$, $60^{\circ}$ above the horizontal. How far above the ground, ball passes over a fielder 50 m from the bat (conside: the ball is struck very close to the ground)? Take $\sqrt{3} = 1.7$ and $g = 10 ms^{-2}$
Explanation
$y = x \tan \theta - (1/2)(gx^{2}/(u^{2}\cos^{2} \theta))$ $y = 50 \tan 60^{\circ} - (10 \times 50 \times 50)/(2 \times 25 \times 25 \times \cos^{2} 60^{\circ}) = 5 m$