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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

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

Can you answer this question?

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
edited Jan 25, 2014 by meena.p

+1 vote

+1 vote