A particle is projected with velocity 2 square root of gh at such an angle that it just clears two walls of equal height h which are at a distance 2h from each other. Show that time of passing between the two walls is 2 square root of straight h over straight g end root.


Let the particle be projected at angle θ. The time taken by particle to cross the two walls each of height h is equal to difference of instants when it would be at a height h above ground. The kinematics of motion of particle in vertical direction is,

Let the particle be projected at angle θ. The time taken by particle

Let the particle be projected at angle θ. The time taken by particle

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A body is projected with velocity u from the top of building h meter high. Find the angle of projection so that it falls at farthest point.

Let the stone be projected at angle θ to make it fall at farthest point R from the foot of tower.

Let the stone be projected at angle θ to make it fall at farthest po

Let the stone be projected at angle θ to make it fall at farthest po

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Two hunters simultaneously aim their guns at a bird sitting on a tree. The first hunter fires his shot with velocity 117·6m/s at angle 30°. The second hunter ahead of first hunter by a distance 19·6√3 m fires his shot at angle 60° and both the shots hit the bird simultaneously. Find the speed of fire of shot by second hunter, the distance of foot of tree from first hunter and height of bird.

The situation is as shown in figure. Let w.r.t. A (point of fire of first shot), the position of bird is straight P left right arrow left parenthesis straight x comma space straight y right parenthesis.

The situation is as shown in figure. Let w.r.t. A (point of fire of f
Position of first shot at any instant:

The situation is as shown in figure. Let w.r.t. A (point of fire of f
As both the shots hit the bird simultaneously, therefore at the instant of hitting positions of both the shots will be same.
i.e.                  straight x subscript 1 equals straight x subscript 2 space space space and space space straight y subscript 1 equals straight y subscript 2
From condition straight y subscript 1 equals straight y subscript 2 comma we get

The situation is as shown in figure. Let w.r.t. A (point of fire of f

              

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A stone is projected with velocity (√2gh) from the top of building h meter high. Show that it will fall farthest at a distance (2√2)h.


Let the stone be projected at angle θ to make it fall at farthest point and (0,R) be the coordinates of point where stone reaches the ground.


Let the stone be projected at angle θ to make it fall at farthest po


Let the stone be projected at angle θ to make it fall at farthest po
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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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Therefore, the body falls at a distance of 200square root of 3 from the point of projection. 
      

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