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2014-10-28T03:27:56+05:30

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Let us say the the project is sent in to air at an angle of Ф with the horizontal and with an initial speed of u in that direction.

There is a deceleration in the vertical direction = -g.

The speed of projectile in the vertical direction is :  v = u + at =>
                           Vy = u sin Ф - g t        --  equation 1

The speed in the horizontal direction is :   Vx = u cos Ф.     --  equation 2
There is no acceleration in X-direction.

The distance travelled is :        s = ut+1/2 at²  

          =>    Sx = x = u (cos Ф) t         --- equation  3
                      t = x / (u cos Ф)

            Sy = y = (u Sin Ф)  t  -  1/2 g t²

            y =  u sinФ x / u cosФ - 1/2 g x² / u² Cos² Ф
 
                 y = x tan Ф - g x² / (2u²Cos²Ф)           ---   equation 4

====================================

y = H when  Vy becomes 0.   =>  Vy = 0 = u sin Ф - g t   
         =>  t  =  u sin Ф / g 
  y = u Sin Ф * (u sin Ф / g)  - 1/2 g (u SinФ/g)²

  y = H =  u² Sin² Ф / 2 g    --   equation  5

  y = 0 =>  x = R
  
      Substituting in equation 4, we get   R = u² Sin 2Ф / g       --  equation 6

1 5 1