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A body is projected from the earth at angle $30^{\circ}$ with the horizontal with some initial velocity. If its range is $20 \mathrm{~m}$, the maximum height reached by it is : (in metres)
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Verified Answer
The correct answer is:
$\frac{5}{\sqrt{3}}$
$R=\frac{u^2 \sin 2 \theta}{g}$
$\therefore \quad 20=\frac{u^2 \sin \left(2 \times 30^{\circ}\right)}{g}$
$\Rightarrow \quad \frac{u^2}{g}=\frac{20}{\sin 60^{\circ}}=\frac{20}{\sqrt{3}} \times 2=\frac{40}{\sqrt{3}}$
Now, $H=\frac{u^2 \sin ^2 \theta}{2 g}$
$=\frac{40}{\sqrt{3}} \times \frac{\sin ^2 30^{\circ}}{2}$
$=\frac{40}{\sqrt{3}} \times \frac{\left(\frac{1}{2}\right)^2}{2}$
$=\frac{5}{\sqrt{3}} \mathrm{~m}$
$\therefore \quad 20=\frac{u^2 \sin \left(2 \times 30^{\circ}\right)}{g}$
$\Rightarrow \quad \frac{u^2}{g}=\frac{20}{\sin 60^{\circ}}=\frac{20}{\sqrt{3}} \times 2=\frac{40}{\sqrt{3}}$
Now, $H=\frac{u^2 \sin ^2 \theta}{2 g}$
$=\frac{40}{\sqrt{3}} \times \frac{\sin ^2 30^{\circ}}{2}$
$=\frac{40}{\sqrt{3}} \times \frac{\left(\frac{1}{2}\right)^2}{2}$
$=\frac{5}{\sqrt{3}} \mathrm{~m}$
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