A particle projected with a velocity u strikes at right angles a plane through the point of projection inclined at an angle 
to the horizon
a) Find the height of the point struck above the horizontal plane through the point of projection is?
b)The time of the flight t upto that instant is?
How do we do this one?
Taking into account that the projectile hits the inclined plane at right angles, i got the relation
)
Now how do we proceed??
Moderation Team
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![\mbox{Considering X and Y axes as shown } \\ \\<br/>Final \ horizontal \ velocity=v_x=u \cos (\alpha - \beta) - g sin \beta t \\ \\ =0 \mbox{ As it falls perpendicular to the plane} \\ \\<br/>Thus , t=\frac{u \cos (\alpha - \beta)}{g \sin \beta} \\ \\<br/>R=\mbox{Range on inclined plane } \\ \\<br/>=u_xt +\frac{at^2}{2} \\ \\<br/>=u \cos (\alpha - \beta) \frac{u \cos (\alpha - \beta)}{g \sin \beta} - g \sin \beta \frac{[\frac{u \cos (\alpha - \beta)}{g \sin \beta}]^2}{2} \\ \\<br/>=\frac{\frac{[u \cos (\alpha - \beta)]^2}{g \sin \beta}}{2} \\ \\<br/>Also , \sin \beta=\frac{h}{R} \\ \\<br/>Hence, \\ \\<br/>\mboxed{h=\frac{[u \cos (\alpha - \beta)]^2}{2g}}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/2/d/9/2d91faeb37ef5bc470608e546b90e0515157c3a2.gif)
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