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Prove that projectile motion is parabola?

Answer» Path of a projectilelet OX be a horizontal line on the ground and OY be a verticle line O is the origin for X and Y axis Consider that the projectile is fired with velocity u and making angle θ with the horizontal form the point ′O′ on the groundThe velocity of projection of the projectile can be resolved into the following two components\xa0(i) ux\xa0= ucosθ, alongOX(ii) uy\xa0= usinθ, alongOY\xa0As the projectile moves it covers distance along the horizontal due to the horizontal component\xa0ucosθ\xa0of the velocity of projection and along verticle due to the vertical component\xa0usinθ\xa0. let that any time t , the projectile reaches the point P , So that its distance along the\xa0X\xa0and\xa0Y\xa0axis are given by x and y respetively.Motion along horizontal direction : it we neglect friction due to air, then horizontal component of the velocity i.e. ucosθ will remain constant. thusInitial velocity along the horizontal ux\xa0= ucosθAcceleration along the horizontal ax = 0The position of the projectile along X-axis at any time t is given byx = uxt + {tex}\\frac 12{/tex}axt2Putting\xa0ux \u200b= ucosθ\xa0and\xa0ax = 0\xa0we have\xa0x = (u cos θ)t + {tex}\\frac 12{/tex}\u200b(o)t2or\xa0x = (u cos θ)tor\xa0t = {tex}\\frac x{u \\cos \\theta}{/tex}\xa0\u200b....... (i)Motion along verticle direction:\xa0The velocity of the projectile along the verticle goes on decreasing due to effect of gravity initial velocity along vertical\xa0uy\xa0= usinθAcceleration along vertical\xa0ay=−gThe position of the projectile along T-axis at any time t is given by\xa0y = uyt + {tex}\\frac 12{/tex}ayt2Putting\xa0uy\u200b=usinθ\xa0and\xa0ay\u200b=−g\xa0we have\xa0y = (u sin θ)t +\xa0{tex}\\frac 12{/tex}(−g)t2or\xa0y = (u sin θ)t −\xa0{tex}\\frac 12{/tex}gt2.....(2)Putting the value of t from equation (1) in equation (2) we get\xa0y = (u sin θ) {tex}\\frac x{u \\cos \\theta}{/tex} − {tex}\\frac 12{/tex}\xa0\u200bg({tex}\\frac x{u \\cos \\theta}{/tex})2or\xa0y = x tan θ − ({tex}\\frac g{2\\ u^2\\ cos^2\\theta}{/tex})x2...(3)This is an equation of a parabola . hence the path of projectile projected at some angle with the horizontal direction is a parabola.


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