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 ?

Explanation

Let the particle projected from (O) strikenthe inclinde plane (O A) at (P) after time (t) and coordinates of (P) be (x, y) Fig. 2 (d) .28.
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Taking motion fo projectile from ` o to P` along x-axis we have `x_(0) =0, x=x, u_(x) =u, a_(x) =0, t=t`.
Using the relation, `x =x_(0) + u_(x) t + 1/2 ax_(x) t^(2)`,
we get `x= =ut or t= x//u` ..(i)
Taking motion of projectile along Y-axis, have `y_(0) =0, y =y, u_(y) =0, a_(y) 0, a_(y) =g, t=t`.
Uaing the relation, `y=y_(0) + u_(y) t + 1/2 a_(y) t^(2)`,
we have `y=0+0 +1/2 at^(2)`
`= 1/2 gt^(2) =1/2 g x^(2)/u^(2) [From (i)]`
Here, `y= x tan theta, so ( g x^(2)0/(2 u^(2)) =x tan theta`
or `x= (2 u^(2) tan theta)/g `
and `y=x tan theta = (2 u^(2) tan^(2) theta)/g`
:. Distance `OP =sqrt( x^(2) =y^(2))`
`=( 2 u^(2) tan theta )/g sqrt (1 + tan^(2) theta ) =(2 u ^(2) tan theta sec theta)/g`.