A man standing on a hill top projects a stone horizontally with speed v0 as shown in fig. (2). Taking the coordinate system as shown in figure, thecoordinates of the point where the stone will hit the hill surface will be

# A man standing on a hill top projects a stone horizontally with speed v0 as shown in fig. (2). Taking the coordinate system as shown in figure, thecoordinates of the point where the stone will hit the hill surface will be

1. A

$\left(\frac{2{\mathrm{v}}_{0}^{2}\mathrm{tan}\mathrm{\theta }}{\mathrm{g}},-\frac{2{\mathrm{v}}_{0}^{2}{\mathrm{tan}}^{2}\mathrm{\theta }}{\mathrm{g}}\right)$

2. B

$\left(\frac{2{\mathrm{v}}_{0}^{2}}{\mathrm{g}},\frac{2{\mathrm{v}}_{0}^{2}{\mathrm{tan}}^{2}\mathrm{\theta }}{\mathrm{g}}\right)$

3. C

$\left(\frac{2{{\mathrm{v}}_{0}}^{2}\mathrm{tan}\mathrm{\theta }}{\mathrm{g}},\frac{2{{\mathrm{v}}_{0}}^{2}}{\mathrm{g}}\right)$

4. D

$\left(\frac{2{\mathrm{v}}_{0}^{2}{\mathrm{tan}}^{2}\mathrm{\theta }}{\mathrm{g}},\frac{2{\mathrm{v}}_{0}^{2}\mathrm{tan}\mathrm{\theta }}{\mathrm{g}}\right)$

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### Solution:

See fig.

The range of the projectile on an inclined plane is given by

$\mathrm{R}=\frac{{\mathrm{u}}^{2}}{{\mathrm{gcos}}^{2}\mathrm{\beta }}\left[\mathrm{sin}\left(2\mathrm{\alpha }+\mathrm{\beta }\right)+\mathrm{sin}\mathrm{\beta }\right]$    (down the plane)

Here,

Now      $\mathrm{X}=\mathrm{Rcos}\mathrm{\theta }=\frac{2{\mathrm{v}}_{0}^{2}\mathrm{tan}\mathrm{\theta }}{\mathrm{g}}$

and        $\mathrm{y}=-\mathrm{Rsin}\mathrm{\theta }=-\frac{2{\mathrm{v}}_{0}^{2}{\mathrm{tan}}^{2}\mathrm{\theta }}{\mathrm{g}}$

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