Question

A pariicle is projected horizontally with a speed (u) from top of a plance inclined at anangle `theta` with the horizontal direction. How far from the point of projection will the particle strike the plane ?

Answer 1

Suppose that the particle strikes the plane at a point `P` with corrdinates `(x,y)`. Consider the motion between points `A` and `P` by resolving them into two directions
Motion in `x`-direction
Initial velocity `= u`,
acceleration `= 0`
`:. x = ut ...(i)`
Motion in `y`-direction
Initial velocity `=0`,
acceleration `= g`,
`y =(½) "gt"^(2) ...(ii)`
Eliminating `t` from (i) & (ii)
`y = (1)/(2)g (x^(2))/(u^(2))`
Also, `y = x tan theta`
Therefore `(gx^(2))/(2u^(2)) = xtan theta rArr (gx^(2))/(2u^(2)) -x tan theta = 0`
`rArrx ((gx)/(2u^(2))-tan theta) =0 rArr x(2u^(2) tan theta)/(g) ...(iii)`
Thus, `y = x tan theta = (2u^(2) tan^(2)theta)/(g) ...(iv)`
`:.` Distance `AP = sqrt(x^(2)+y^(2)) = (2u^(2))/(g) tan theta`
`= sqrt((1+tan^(2)theta)) = (2u^(2))/(g) tan theta sec theta`,