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 ?

Explanation

Take X,Y-axis as shown in ure Suppose that the particle strikes the plane at a point P with coordiN/Ates (x,y). Consider the motion between A and P.

Motion in x- direction
`Initial velocity = u`
`Acceleration =g`
`y=1/2 gt^2`
ElimiN/Ating t from i. and ii
`y= 1/2 g x^2/u^2`
Also `y=xtantheta`
`Thus, `(gx^2)/(2u^2)=x tan theta `giving `x=0, or, `(2u^2tantheta)/g,
then `y= x tantheta= (2u^2 tan^2 theta)/g`
The distance `AP=l=sqrt(x^2+y^2)`
`=(2u^2)/g tan theta sqrt (1+tan^2theta)`
`=(2u^2)/g tantheta sec theta.` AlterN/Atively: Take the axes as shown in ure. consider the motion between A and P.

Motion along the X-axis:
Intial velocity `=u cos theta`
Acceleration `=g sin theta`
`Displacement =AP`.
Thus, `AP=(u cos theta)t+1/2(g sin theta)t^2`
Motion along the Y-axis
`Initial velocity =-u sin theta`
`acceleration =g cos theta`
`Displacement =0`
Thus, ` 0=- uy sin theta +1/2 gt^2 cos theta`
`or, `t=0, (2u sintheta)/(g(costheta)`
clearly the point P corresponds to `t=(2u sin theta)/(g cos theta)`
Putting this value of t in i.
`AP=(u cos theta)((2u sin theta)/(g costheta))+(g sin theta)/2((2u sin theta)/(g costheta))^2`
`(2u^2 sin theta)/g+(2u^2 sintheta tan^2 theta)/g`
`=(2u^2)/g sin theta sec^2 theta=(2u^2)/g tantheta sec theta`.