Question:
A ball is thrown at a speed of $40 \mathrm{~m} / \mathrm{s}$ at an angle of $60^{\circ}$ with the horizontal. Find
(a) The maximum height reached
(b) The range of the ball
Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$
Solution:
(a) $a x=\frac{\mathrm{u}^{2} \sin 2 \theta}{2 \mathrm{~g}}=\frac{40^{2}\left(\sin ^{2} 60\right)}{2 \mathrm{~g}}$
$\mathrm{H}_{\max }=60 \mathrm{~m}$
(b) $R=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}=\frac{40^{\mathrm{a}} \sin (2 \mathrm{x} 60)}{\mathrm{g}}$
$R=80 \sqrt{3} m$
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