Question

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

• A. $\frac{2v^{2}}{g}$
• B. $\sqrt{2}\left[\frac{2v^{2}}{g}\right]$
• C. $\frac{v^{2}}{g}$
• D. $\sqrt{2}\frac{v^{2}}{g}$

Answer 1

Answer: (B)

$x= vt\,$ or $\, y=\frac{1}{2} gt^{2}$

$or \,\, y=\frac{gx^{2}}{2v^{2}} \,\,or\
\ =\frac{gx^{2}}{2v^{2}} \,\,or\,\, x=\frac{2v^{2}}{g}$
Also, $\,\,y=\frac{2v^{2}}{g} \,\,or\,\, l^{2}=x^{2}+y^{2}$
$l=\sqrt{2} \times x$
or $l=\sqrt{2}[\frac{2v^{2}}{g}]$