Question

show that path of the projectile is parabola making an angle tita with the horizontal plane..??

Answer 1

1.Moving along horizontal - If we neglect friction due to air ,then horizontal component of the velocity i.e. ucos theta will remain constant.thus,initial velocity along the horizontal,

                                                            ux=u cos theta

  acceleration along the horizontal,

                                                            ax=0

The position of the projectile along X-axis at any time t is given by

                                          x= ux +1/2 at2

setting uy = u sin theta and ax =0,we have

                                        x=(u cos theta )t + 1/2 at2

       OR                              t = x/u cos theta                                        ...............(i)

 

2. Motion along vertical - The velocity of the projectile along the vertical goes on decreasing  due to the effect of gravity.

Initial velocity along vertical,uy= u sin theta 

acceleration along vertical,ay=  --g

The position of the projectile along Y-axis at any time t is given by

                                                   y= uy t +1/2ayt2

setting uy=u sin theta and ay=--g,we have 

                                                   y== (u sin theta )t + 1/2(--g)t2 

                                                  y = (u sin theta )t -- 1/2 gt2                                  ............. (ii)

(i)Equation of  trajectory . the equation of the trajectory of the projectile can be obtained by substituting the value of t from equation (i) in the equation (ii).Therefore,we have 

                                                y=(u sin theta ) x/u cos theta  -- 1/2 g(x/u cos theta)2 

    OR                                      y= x tan theta --(g/2u2 cos2)2                                  ............(iii) 

As equation (iii) is the equation of parabola,it follows that a projectile fired at some angle theta  with the horizontal moves along a parabolic path