A particle is projected horizontally with a speed ''u'' rom 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 ?

Explanation

Suppose particle is projected from height y , it strikes the ground at a horizotnal distance x , then distance `R = sqrt(x^(2) + y^(2))" "……(i)`
If "t" is the time of motion , then
x = ut
and `y = (1)/(2) "gt"^(2)`
From equation (i), we get `t = (x)/(u)`
`therefore y = (1)/(2) g((x)/(u))^(2) = (gx^(2))/(2u^(2))" ".........(iii)`
From figure `(y)/(x) = tan theta or y = x tan theta " "...........(iv)` .
From equation (iii) and (iv) , we have
`x tan theta = (gx^(2))/(2u^(2)) or x[(gx)/(2u^(2)) - tan theta] = 0`
As x = 0 is not possible ,
So `[(gx)/(2u^(2)) - tan theta] = 0 or x = (2u^(2) tan theta)/(g)" ".......(v)`
Now form equaiton (i) , (iv) and (v) we , get .
`R= sqrt(x^(2)+ y^(2)) = sqrt(x^(2) + (x tan theta)^(2)) = x sqrt(1+tan^(2) theta) = s sec theta or = (2u^(2))/(g) tan theta sec theta`