A ball is projected horizontally with a speed V from the top of a plane inclined at angle $45^{0}$ with the horizontal. How far from the point of projection will the ball strike the plane?

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$\frac{V^{2}}{g}$

$\sqrt{2} \frac{V^{2}}{g}$

$\frac{2 V^{2}}{g}$

$\frac{2 \sqrt{2} V^{2}}{g}$

answer is D.

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Detailed Solution

$\begin{array}{l} down the plane Range =R = \frac{u^{2}}{g \cos^{2} β} [\sin (2 α + β) + \sin β] \\ h e r e α = 0 = a n g l e o f p r o j e c t i o n i s a l o n g h o r i z o n t a l l i n e; s u b s t i t u t e \\ R = \frac{u^{2}}{g \cos^{2} 45^{\circ}} (\sin 45^{\circ} + \sin 45^{\circ}) \\ R = \frac{u^{2}}{g (\frac{1}{2})} \times 2 \times (\frac{1}{\sqrt{2}}) \\ R = 2 \sqrt{2} \frac{V^{2}}{g} [Here u = v] \end{array}$