Define a projectile and derive the equation of its trajectory: $y = (\tan \theta_0)x - \frac{g}{2(v_0 \cos \theta_0)^2}x^2$.

(N/A) Definition: An object projected into the air,which then moves under the influence of gravity alone,is called a projectile,and its motion is called projectile motion.
Derivation:
Let an object be projected with an initial velocity $v_0$ at an angle $\theta_0$ with the horizontal. The components of initial velocity are $v_{0x} = v_0 \cos \theta_0$ and $v_{0y} = v_0 \sin \theta_0$.
The horizontal position at time $t$ is given by:
$x = (v_0 \cos \theta_0)t \implies t = \frac{x}{v_0 \cos \theta_0} \quad \dots(1)$
The vertical position at time $t$ is given by:
$y = (v_0 \sin \theta_0)t - \frac{1}{2}gt^2 \quad \dots(2)$
Substituting the value of $t$ from equation $(1)$ into equation $(2)$:
$y = (v_0 \sin \theta_0) \left( \frac{x}{v_0 \cos \theta_0} \right) - \frac{1}{2}g \left( \frac{x}{v_0 \cos \theta_0} \right)^2$
Simplifying the expression:
$y = x \tan \theta_0 - \frac{g}{2(v_0 \cos \theta_0)^2}x^2$
This is the equation of the trajectory of a projectile.