$A$ 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).

  • A
    $\frac{4 v^2}{5 g}$
  • B
    $\frac{4 g}{5 v^2}$
  • C
    $\frac{v^2}{g}$
  • D
    $\frac{4 v^2}{\sqrt{5} g}$

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Motion in two dimensions in a plane can be studied by expressing position,velocity,and acceleration as vectors in Cartesian coordinates $\vec{A} = A_{x} \hat{i} + A_{y} \hat{j}$,where $\hat{i}$ and $\hat{j}$ are unit vectors along $x$ and $y$ directions,respectively,and $A_{x}$ and $A_{y}$ are corresponding components of $\vec{A}$. Motion can also be studied by expressing vectors in circular polar coordinates as $\vec{A} = A_{r} \hat{r} + A_{\theta} \hat{\theta}$,where $\hat{r} = \cos \theta \hat{i} + \sin \theta \hat{j}$ and $\hat{\theta} = -\sin \theta \hat{i} + \cos \theta \hat{j}$ are unit vectors along the directions in which $r$ and $\theta$ are increasing.
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$(c)$ Show that $\frac{d}{dt}(\hat{r}) = \omega \hat{\theta}$,where $\omega = \frac{d\theta}{dt}$ and $\frac{d}{dt}(\hat{\theta}) = -\omega \hat{r}$.
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