Question

What is the Motion of Centre of mass of a projectile fired at an angle X with a velocity V.

Answer 1

I am changing symbols for speed and angle.

     Let us say the the projectile is fired into air at an angle of Ф with the horizontal and with an initial speed of u in that direction.  Initially, at time t = 0sec, the projectile is at origin O(0,0) at height =0.  At time t sec, the projectile is on the parabolic path at a point P (x, y) with instantaneous velocities Vx, and Vy in x and y directions.

There is a deceleration in the vertical direction = - g

  Speed of projectile in the vertical direction is :  V = u + a * t

         =>  Vy = u sin Ф - g t        --  equation 1

Speed in the horizontal direction is a constant as there is no acceleration: 

       =>    Vx = u cos Ф + 0 * t      --  equation 2

    There is no acceleration in X-direction.

Displacement = Distance traveled is :  s = u t + 1/2 a t²  

      =>    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 sin Ф) (x / u cos Ф) - 1/2 g (x² / u² Cos² Ф)
 
                 y = x tan Ф - g x² / (2u²Cos²Ф)           ---   equation 4

   This is the equation of motion of the projectile, in 2 dimensions.  This is the locus.

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The maximum height reached by projectile is H.

  y = H when  Vy becomes 0.   =>  Vy = 0 = u sin Ф - g t   
         =>  t  =  u sin Ф / g  = time interval to reach the maximum height.

    H = u Sin Ф * (u sin Ф / g)  - 1/2 g (u SinФ/g)²

  Hence,  H =  u² Sin² Ф / 2 g    --   equation  5

    y = 0 =>  x = R = Range of the projectile , where it lands on ground.
  
      Substituting in equation 4, we get

           R = u² Sin 2Ф / g       --  equation 6