When an object is projected at an angle to the horizontal, its motion can be analyzed by separating it into horizontal (x) and vertical (y) components. The horizontal motion is characterized by constant velocity ($v_x = v_0 \cos heta$), as there are no horizontal forces acting on the projectile (ignoring air resistance). The vertical motion, however, is influenced by gravity, resulting in constant downward acceleration ($a_y = -g$). The initial vertical velocity is $v_{0y} = v_0 \sin heta$.

1. Time of Flight (T): The total time the projectile spends in the air. This occurs when the vertical displacement $y(t)$ returns to zero (assuming launch and landing at the same height). Setting $y(T) = 0$ and solving for $T$ gives $T = \frac{2v_0 \sin \theta}{g}$.
2. Maximum Height (H): The highest vertical position reached by the projectile. This occurs when the vertical velocity $v_y(t)$ becomes zero. Setting $v_y(t_{peak}) = 0$ and solving for $t_{peak}$ gives $t_{peak} = \frac{v_0 \sin \theta}{g}$. Substituting this time into the $y(t)$ equation gives $H = \frac{(v_0 \sin \theta)^2}{2g}$.
3. Horizontal Range (R): The total horizontal distance covered by the projectile. This is the horizontal displacement $x(t)$ at the total time of flight $T$. $R = x(T) = (v_0 \cos \theta) T = \frac{2v_0^2 \sin \theta \cos \theta}{g} = \frac{v_0^2 \sin(2\theta)}{g}$. The range is maximum when $\sin(2\theta)$ is maximum, which occurs at $2\theta = 90^\circ$, or $\theta = 45^\circ$.

From the horizontal displacement equation, $t = \frac{x}{v_0 \cos \theta}$. Substituting this into the vertical displacement equation $y(t) = (v_0 \sin \theta) t - \frac{1}{2}gt^2$, we get: $y = (v_0 \sin \theta) \left(\frac{x}{v_0 \cos \theta}\right) - \frac{1}{2}g \left(\frac{x}{v_0 \cos \theta}\right)^2$ $y = x \tan \theta - \frac{gx^2}{2v_0^2 \cos^2 \theta}$ This is the equation of the trajectory, which is a parabola.