Circular motion, projectile motion Why does a body thrown upward with certain velocity, at some angle from vertical, travel on a parabolic path but not a circular path? 
Since there is always a component of weight along the center, it should follow a circular path.
 A: Say the projectile was thrown with a velocity $v$ at an angle $\theta$ with respect to the horizontal. We ignore all friction effects (air drag, side winds). Define a coordinate system with a vertical $y$-axis, a horizontal $x$-axis and the point of origin $O$ the point from which the projectile starts its trajectory.
The trajectory can now be decomposed in a vertical and a horizontal component.
The horizontal component is $x=(v \cos \theta) t$.
The vertical component is subject to gravitational deceleration, so we can write the vertical velocity component as $\frac{dy}{dt}=v \sin \theta -gt$.
So that $dy=(v \sin \theta -gt)dt$.
Integration yields: $y=(v \sin \theta)t-\frac{1}{2} gt^2 + C$.
The boundary condition $t=0, y=0$ tells us that $C=0$, so:
$y=(v \sin \theta)t-\frac{1}{2} gt^2$.
From the vertical component $x=(v \cos \theta) t$ we isolate $t$:
$t = \frac{x}{v \cos \theta}$.
Substituting into $y$ gives:
$y=(\tan \theta) x-\frac{1}{2} \frac{gx^2}{v^2 \cos^2 \theta}$.
This is a second degree polynomial in $x$ and the formula of a parabola.
