# Please Solve Q. D.2.... Answer given is Option (D)...

Ans. to Q. D-2

$Forahorizontalprojectionofabodywithavelocityv\phantom{\rule{0ex}{0ex}}h=x=\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}x=horizontalrange=vt\phantom{\rule{0ex}{0ex}}Theplaneisinclinedat{45}^{0}\phantom{\rule{0ex}{0ex}}\frac{1}{2}g{t}^{2}=vt\phantom{\rule{0ex}{0ex}}t=\frac{2v}{g}\phantom{\rule{0ex}{0ex}}Therefore\phantom{\rule{0ex}{0ex}}x=v\times \frac{2v}{g}=\frac{2{v}^{2}}{g}\phantom{\rule{0ex}{0ex}}wehavetofindAB=\frac{x}{\mathrm{sin}45}=\frac{2\sqrt{2}{v}^{2}}{g}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Regards$

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