A body is projected from the ground with some velocity. Three seconds later the body is moving in a direction making an angle of 30° with the horizontal and two seconds later it moves horizontally. Find the distance at which the body will fall on the ground from the point of projection. from Physics Motion in A Plane Class 11 Manipur Board
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A body is projected from the ground with some velocity. Three seconds later the body is moving in a direction making an angle of 30° with the horizontal and two seconds later it moves horizontally. Find the distance at which the body will fall on the ground from the point of projection.

Let the body be projected with velocity u making an angle θ with the horizontal.

The direction of motion of body after t seconds of throw is given by the equation,

             tan space straight alpha space equals space fraction numerator straight u space sin space straight theta space minus space gt over denominator straight u space cos space straight theta end fraction

Here at time, t = 3 the value of straight alpha is 30 degree and at t = 5s the value of straight alpha is 0 degree.

∴               tan space 30 degree space equals space fraction numerator straight u space sinθ minus 30 over denominator straight u space cosθ end fraction           ...(1)

and  

         tan space 0 degree space equals space fraction numerator straight u space sinθ minus 50 over denominator straight u space cosθ end fraction                       ...(2)

From (2),
 
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Substituting straight u space sinθ space equals 50 in equation (1), we get

              straight u space cos space straight theta space equals space fraction numerator straight u space sinθ minus 30 over denominator tan 30 degree end fraction equals 20 square root of 3      ...(4)

The horizontal range of the body,

             space space space space straight R equals fraction numerator straight u squared sin 2 straight theta over denominator straight g end fraction equals fraction numerator 2 straight u space sinθ space straight u space cosθ over denominator straight g end fraction
                   
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#6 {main}</pre>

Therefore, the body falls at a distance of 200square root of 3 from the point of projection. 
      

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