Projectile under an Angular Projection

Now let the projectile be launched with speed $u$ at an angle $\theta$ to the horizontal (an oblique/angular projection). Resolve the initial velocity into horizontal and vertical components:

Horizontal motion: $x = u_x t = (u\cos\theta)\,t$, so $t=\dfrac{x}{u\cos\theta}$.

Vertical motion: $v_y = u_y - gt = u\sin\theta - gt$, and $y = u_yt-\tfrac12gt^2 = (u\sin\theta)t - \tfrac12gt^2$.

Trajectory equation. Substituting $t=x/(u\cos\theta)$ from the horizontal relation into the vertical-displacement relation eliminates $t$ and gives the shape of the path directly as $y$ vs $x$:

This is once again a parabola in $x$ (an inverted one, opening downward), confirming that both horizontal and oblique projectile paths are parabolic.

Maximum height $h_{max}$: at the highest point of the path the vertical velocity momentarily vanishes, $v_y=0$. Using $v_y^2=u_y^2-2gs$ with $u_y=u\sin\theta$, $s=h_{max}$, and $v_y=0$:

Time of flight $T_f$: the total time from launch until the projectile returns to the same horizontal level it was launched from (net vertical displacement $y=0$). From $y=u_yt-\tfrac12gt^2=0$ (the non-zero root, since $t=0$ is the launch instant itself):

(Notice $T_f = 2\times$ the time to reach the peak — the rise and fall halves take equal time, since gravity is constant and the vertical launch/landing speeds are equal in magnitude.)

Horizontal range $R$: the horizontal distance covered during the full time of flight, $R = u_x \times T_f = (u\cos\theta)\times\dfrac{2u\sin\theta}{g}$, which simplifies (using $2\sin\theta\cos\theta=\sin2\theta$) to

The range depends on the initial speed squared and on $\sin2\theta$, and is inversely proportional to $g$. …