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

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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.

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