Question

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

Answer 1

`{:("x-axis",,"y-axis"),(u_(x)=u,,u_(x)=0),(a_(x)=0,,a_(x)=g),(x=ut,,y=(g t^(2))/2):}`
`implies y=(g x^(2))/(2 u^(2))`
also `y/x=tan theta implies x tan theta=(g x^(2))/(2 u^(2))`
`x=theta, (2u^(2) tan theta)/g`
`x=(2u^(2) tan theta)/g`
`implies y=(2u^(2) tan^(2) theta)/(g)`
dist. `l=sqrt(x^(2)+y^(2))`
`l=(2u^(2) tan theta sec theta)/(g)`

Alternate Method.
`R=(2u^(2) sin (alpha+theta) cos alpha)/(g cos^(2) theta)`
`R=2u^(2) tan theta sec theta`