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) = tan theta)` `= sqrt(x^(2) + (x tan theta)^(2)) = xsqrt(1+ tan^(2) theta) = x SEC theta` `x = u t , y = (1)/(2)g t^(2) , (y)/(x) = (1)/(2)(g t^(2))/(u t)` `tan theta = (g t)/(2u), t = (2u)/(g) tan theta` `x = u t = (2u^(2))/(g) tan theta, ""THEREFORE R = (2u^(2))/(g) tan theta sec theta`