Show that a projectile traces a path of a parabola.

Explanation

Let `v_0` be the velocity of projection. Let `theta` be the angle of projection. Let `v_("ok")=v_0costheta` and `v_0y=v_0sintheta` represent X and Y components of velocities.
For the motion in (x,y) plane
`y=(v_0sintheta)t-1//2g t^2`
and `x=(v_0costheta)t`
i.e `t=((x)/(v_0costheta))`
Substituting for it in (1) in the expression for y we get
`y=(v_0sintheta)((x)/(v_0costheta))-1/2g((x^2)/(v_0^2cos^2theta))`


i.e. `y=xtantheta-x^2(g/(2v_0^2cos^2theta))`
Put `a=tantheta` and b`=((g)/(2v_0^2cos^2theta))`
We get `y-ax-bx^2` which is quadratic equation representing equation of a parabola Hence trajectory of a projectile describes a parabola.