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Motion in A Plane

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Physics Part I

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Physics

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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?

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. 



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