एक कण किसी समतल में नियत त्वरण से किन्तु प

A particle moves with constant acceleration in a direction different from initial

velocity. Let the velocity be $v,$ acceleration be $a,$ angle between them is $\theta\left[\theta<90^{0}\right]$

Component of Velocity in the Direction of $a$ is $v \cos \theta$ and in a direction perpendicular to $a$ is vsin $\theta$

From equations of Kinematics, In the direction of acceleration, $y=v \cos \theta t+\frac{1}{2} a t^{2}-----(1)$

In the direction perpendicular to acceleration,

$x=v \sin \theta t$

$t=\frac{x}{v \sin \theta}--------(2)$

Substituting 2 in $1,$ we get:

$y=v \cos \theta\left(\frac{x}{v \sin \theta}\right)+\frac{1}{2} a\left(\frac{x^{2}}{\sin ^{2} \theta}\right)$

$y=\frac{x}{\tan \theta}+\frac{1}{2} a \frac{x^{2}}{\sin ^{2} \theta}$

This is in the form of a parabola: $y=A x^{2}+B x+C$