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A cricket ball is thrown by a player at a speed of $20 \mathrm{~m} / \mathrm{s}$ in a direction $30^{\circ}$ above the horizontal. The maximum height attained by the ball during its motion is $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$ :
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$5 \mathrm{~m}$
Given : \(\mathrm{u}=20 \mathrm{~m} / \mathrm{s} \theta=30^{\circ}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\)
We get \(\sin \theta=\sin 30^{\circ}=\frac{1}{2}\)
Maximum height reached \(\mathrm{H}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{20 \times 20 \times\left(\frac{1}{2}\right)^2}{20}=5 \mathrm{~m}\)
We get \(\sin \theta=\sin 30^{\circ}=\frac{1}{2}\)
Maximum height reached \(\mathrm{H}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{20 \times 20 \times\left(\frac{1}{2}\right)^2}{20}=5 \mathrm{~m}\)
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