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Motion In A Plane

Question
CBSEENPH11018554

A projectile is projected horizontally with velocity from the top of an inclined plane of inclination θ. How far from the point of projection will the projectile hit tht plane?

Solution
Consider a projectile, which is projected with velocity 'u' in horizontal direction from a point O on an inclined plane and hit the inclined plane at P↔(x, y).
           
Let O be the origin. 
The horizontal direction is positive direction of X-axis and vertically downward direction is positive direction of Y-axis.
On resolving the components of velocity and acceleration in horizontal and vertical direction, we have
space space space space straight x subscript straight o equals 0 space space space space space space space space straight y subscript straight o equals 0
space space space space straight u subscript straight x equals straight u space space space space space space space space straight u subscript straight y space equals 0 space
space space space space straight a subscript straight x equals 0 space space space space space space space space space straight a subscript straight y space equals space straight g 
Motion along horizontal direction is, 
        space straight x equals straight x subscript straight o plus straight u subscript straight x straight t plus 1 half straight a subscript straight x straight t squared equals ut             ...(1) 
Motion along vertical direction is, 
         straight y equals straight y subscript straight o plus straight u subscript straight y straight t plus 1 half straight a subscript straight y straight t squared equals 1 half gt squared      ...(2) 
Eliminating 't' from (1) and (2), 
                straight y equals fraction numerator straight g over denominator 2 straight u squared end fraction straight x squared                        ...(3) 
The equation of inclined plane is, 
                 space straight y equals xtanθ                         ...(4) 
The point P is the intersection of trajectory and inclined plane, thus the coordinates of P are the solutions of equation (3) and (4).
By solving (3) and (4), we obtain 
       space space space straight x equals fraction numerator 2 straight u squared tanθ over denominator straight g end fraction space space space space and space space space space space space straight y space equals space fraction numerator 2 straight u squared tan squared straight theta over denominator straight g end fraction 
Therefore distance OP is, 
          OP equals square root of straight x squared plus straight y squared end root space equals space fraction numerator 2 straight u squared tanθ over denominator straight g end fraction square root of 1 plus tan squared straight theta end root 
           space space equals fraction numerator 2 straight u squared tanθsecθ over denominator straight g end fraction
So, the distance travelled by the projectile is OP.