A particle is projected horizontally with a speed "u" from the top of plane incline at an angle "theta" with the horizontal. How far from the point of projection will the particle strike the plane?

A particle is projected horizontally with a speed "u" from the top of

1.	A particle is projected horizontally with a speed "u" from the top of plane incline at an angle "theta" with the horizontal. How far from the point of projection will the particle strike the plane?
Answer» Solution : `R=SQRT(X^(2)+y^(2)) ((y)/(x)=tantheta)` `=sqrt("x"^(2)+("x"tantheta)^(2))=xsqrt(1+tan^(2)theta)="x"sectheta` `x=ut, y = (1)/(2), (y)/("x")=(1)/(2)("gt"^(2))/(ut)` `tantheta=("gt")/(2u), t=(2u)/(G)tantheta` `"x"=ut=(2u^(2))/(g)tantheta"":. R=(2u^(2))/(g)tanthetasectheta`