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Motion in A Plane

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Physics Part I

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Class 10 Class 12
A stone is thrown from the top of a tower 50m high with speed 30m/s at an angle 30° with horizontal. Find the distance from the foot of tower where the stone will hit the ground.

Given that, 


      u = 30 m/s,       straight theta equals 30 degree

∴ x component of velocity, 

straight u subscript straight x equals 30 space cos space 30 degree space equals space 25.98 space straight m divided by straight s

and  

y component of velocity,   

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Let the stone reaches the ground after time t.

                space space space space space space space 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      ... (1)


We have,

       straight y equals 0 comma space space straight y subscript straight o equals 50 comma space space space straight a subscript straight y equals negative 10 straight m divided by straight s squared

Putting the values in equatuion (1), we have 

           0 equals 50 plus 15 straight t minus 5 straight t squared

i.e.,            straight t equals 5 straight s

The distance from the foot of tower where the stone will hit the ground is,

              s = u x t

                 = 25.98 x 5

                 = 104.9m

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