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A particle is projected with a velocity \( v \) so that its horizontal range twice the greatest height
attained. The horizontal range is
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attained. The horizontal range is
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Verified Answer
The correct answer is:
\( \frac{4 v^{2}}{5 g} \)
Velocity of particle $=v$
Horizontal range $=2 \times$ Greatest height attained
Now, Range $=\frac{v^{2} \sin 2 \theta}{g}$
Height $=\frac{v^{2} \sin ^{2} \theta}{2 g}$
$\Rightarrow \frac{v^{2} \sin 2 \theta}{g}=\frac{2 \times v^{2} \sin ^{2} \theta}{2 g} \Rightarrow \sin 2 \theta=\sin ^{2} \theta$
Using, $\sin 2 \theta=2 \sin \theta \cos \theta$, we get
$2 \sin \theta \cos \theta=\sin ^{2} \theta \Rightarrow 2 \cos \theta=\sin \theta \Rightarrow \frac{\sin \theta}{\cos \theta}=2$
$\Rightarrow$ tan $\theta=2$
$\Rightarrow \sin \theta=\frac{2}{\sqrt{5}}$ and $\cos \theta=\frac{1}{\sqrt{5}}$
Therefore,
Range
$=\frac{v^{2} \sin 2 \theta}{g}=\frac{2 v^{2} \sin \theta \cos \theta}{g}=\frac{2 v^{2}}{g} \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}}=\frac{4}{5} \frac{v^{2}}{g}$
$\Rightarrow$ Horizontal range $=\frac{4}{5} \frac{v^{2}}{g}$
Horizontal range $=2 \times$ Greatest height attained
Now, Range $=\frac{v^{2} \sin 2 \theta}{g}$
Height $=\frac{v^{2} \sin ^{2} \theta}{2 g}$
$\Rightarrow \frac{v^{2} \sin 2 \theta}{g}=\frac{2 \times v^{2} \sin ^{2} \theta}{2 g} \Rightarrow \sin 2 \theta=\sin ^{2} \theta$
Using, $\sin 2 \theta=2 \sin \theta \cos \theta$, we get
$2 \sin \theta \cos \theta=\sin ^{2} \theta \Rightarrow 2 \cos \theta=\sin \theta \Rightarrow \frac{\sin \theta}{\cos \theta}=2$
$\Rightarrow$ tan $\theta=2$
$\Rightarrow \sin \theta=\frac{2}{\sqrt{5}}$ and $\cos \theta=\frac{1}{\sqrt{5}}$
Therefore,
Range
$=\frac{v^{2} \sin 2 \theta}{g}=\frac{2 v^{2} \sin \theta \cos \theta}{g}=\frac{2 v^{2}}{g} \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}}=\frac{4}{5} \frac{v^{2}}{g}$
$\Rightarrow$ Horizontal range $=\frac{4}{5} \frac{v^{2}}{g}$
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