A particle is projected horizontal 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?

Explanation

Let us select point projection as the origin and X- and Y-axis as shown in the figure .

There is no acceleration in the horizontal direction hence we can write the following :
`x="ut"rArrt(x)/(u)`
We can use `S="ut"+(1)/(2)"at"^(2)` in the Y-direction . Initial
Velocity along Y-axis is zero and acceleration a=g, as we have selected downward direction as the positive direction .
`S="ut" +(1)/(2)"at"^(2)`
`rArr y=(1)/(2)"gt"^(2)`
Substituting t from equation (i) ,
`rArr y=(1)/(2)g((x)/(u))^(2)rArry=(gx^(2))/(2u^(2))`
We can also write the following equation:
`y=x tan theta`
Substituting from equation (iii) in equation (ii) , we get the following :
`x tan theta=(gx^(2))/(2u^(2))`
`rArr x=(2u^(2))/(g) tan theta`
Substituting in equation (iii) , we get the following :
`y=(2u^(2))/(g) tan^(2) theta`
Distance of the stone point of projection can be written as followings:
`l=sqrt(x^(2)+y^(2))=(2u^(2))/(g) tan theta sqrt(1+tan^(2) thea)`
`rArr l=(2u^(2))/(g)tan theta sec theta`