Question

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

Answer 1

`Rsqrt(x^(2)+y^(2)) (y/x=tan theta)`
`=sqrt(x^(2)+(x tan theta)^(2))=xsqrt(1+tan^(2)theta)=x sec theta`
`x=ut,y=1/2"gt"^(2),y/x=1/2,"gt"^(2)/ut`
`tantheta="gt"/2u, t=(2u)/(g)tantheta`
`x=ut=(2u)/(g)tantheta,`
`:.R=(2u^(2))/(g)tanthetasectheta`