A man standing on a hill top projects a stone horizontally with speed $v_{0}$ as shown in figure. Taking the co.ordinate system as given in the figure. Find the co-ordinates of the point where the stone will hit the hill surface.

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$\frac{3 v_{0}^{2} \tan θ}{g} a n d - \frac{2 v_{0}^{2} \tan^{2} θ}{g}$

$\frac{2 v_{0}^{2} \tan θ}{g} a n d - \frac{3 v_{0}^{2} \tan^{2} θ}{g}$

$\frac{2 v_{0}^{2} \tan θ}{g} a n d - \frac{2 v_{0}^{2} \tan^{2} θ}{g}$

$\frac{2 v_{0}^{2} \tan θ}{g} a n d \frac{2 v_{0}^{2} \tan^{2} θ}{g}$

answer is C.

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Detailed Solution

Range of a projectile projected down an inclined plane

$R = \frac{u^{2}}{g \cos^{2} β} [\sin (2 α + β) + \sin β] Here, u = v_{0}, α = 0 (angle of v_{0} with horizontal) and β = θ (inclination angle of plane)$

$\Rightarrow R = \frac{2 v_{0}^{2} \sin θ}{g \cos^{2} θ} ∴ x = R \cos θ = \frac{2 v_{0}^{2} \tan θ}{g} a n d y = - R \sin θ = - \frac{2 v_{0}^{2} \tan^{2} θ}{g}$