A particle is projected with a velocity \(v\) such that its range on the horizontal plane is twice the…

A particle is projected with a velocity \(v\) such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (g= acceleration due to gravity)
  1. \(\frac{4 v^2}{5 g}\)
  2. \(\frac{4 g}{5 v^2}\)
  3. \(\frac{v^2}{g}\)
  4. \(\frac{4 v^2}{\sqrt{5} g}\)

Solution

Velocity of particle $=v$ If $\theta$ is angle of projection such a way, $\begin{aligned} & R=2 H \\ \Rightarrow & \frac{v^{2} \sin 2 \theta}{g}=\frac{2 \cdot v^{2} \sin^{2} \theta}{2 g} \\ & 2 \sin \theta \cos \theta=\sin^{2} \theta \\ \Rightarrow \quad & 2=\frac{\sin \theta}{\cos \theta} \Rightarrow 2=\tan \theta \Rightarrow \tan \theta=2 \end{aligned}$ $\begin{aligned} \therefore \quad & \sin \theta=\frac{2}{\sqrt{5}} \Rightarrow \cos \theta=\frac{1}{\sqrt{5}} \\ \therefore \quad \text { Range } & =\frac{v^{2} \sin 2 \theta}{g}=\frac{v^{2} \cdot 2 \sin \theta \cos \theta}{g} \\ & =\frac{v^{2} \times 2 \times \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}}}{g}=\frac{4 v^{2}}{5 g} \end{aligned}$

Asked in: AP EAMCET 2020 (17 Sep Shift 1)

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