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A ball is projected horizontaly with a speed v from top of a plane inclined at an angle $45^{\circ}$ with the horizontal. How far from the point of projection will the ball strike the plane

\[(a)\; \frac{v^2}{g}\quad (b)\;j_2\bigg (\frac{v^2}{g}\bigg) \quad (c)\;\frac{2v^2}{g} \quad (d)\sqrt 2 \bigg[\frac{2v^2}{g}\bigg]\]

1 Answer

For a horizontal projection of a body with velocity v
$h=\large\frac{1}{2} $$gt^2$
and $x=horizondal\;range=vt$
Since the plane is inclined at $45^{\circ}$
we have $h=x$
$\large\frac{1}{2} $$gt^2=vt$
$\qquad t= \large \frac{2v}{g}$
$\qquad x=v \times \large\frac{2v}{g}=\frac{2v^2}{g}$
Now $ AB=\large\frac{x}{\sin \theta}=\frac{2v^2}{g} \times \frac{1}{\sin \theta}$
$\qquad=\large\frac{2v^2}{g} $$\sqrt 2$
Hence d is the correct naswer.


answered Jul 4, 2013 by meena.p
edited Jan 25, 2014 by meena.p

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