What launch angle maximizes projectile range on flat ground?

Ignoring air resistance, a 45° launch angle produces the maximum horizontal distance (range) when the launch height equals the landing elevation.

🚀 Classical Kinematics & Physics

Projectile Motion & Trajectory Calculator

Solve 2D parabolic kinematics: initial velocity vectors, apex altitude, flight duration, and maximum horizontal landing distance.

Initial Velocity ($v_0$ in m/s):

Launch Angle ($\theta$ in degrees):

Initial Elevation / Height ($h_0$ in meters):

Gravitational Acceleration ($g$ in m/s²):

Max Range (45°) High Apex (60°) Cliff Drop (0°, 50m)

Maximum Horizontal Range ($R$)
63.73 m
Total Flight Time: 3.60 seconds

Apex Height ($H$)

15.93 m

Horiz. Velocity ($v_x$)

17.68 m/s

Vert. Velocity ($v_{0y}$)

17.68 m/s

Impact Velocity

25.00 m/s

Impact Angle

-45.0°

$$v_x = v_0 \cos\theta, \quad v_y(t) = v_0 \sin\theta - gt, \quad y(t) = h_0 + v_0 \sin\theta \cdot t - \frac{1}{2}gt^2$$

In classical Newtonian mechanics, horizontal and vertical motions are completely independent:

Horizontal Motion: $a_x = 0$, meaning horizontal velocity remains constant throughout flight: $x(t) = v_{0x} \cdot t$.
Vertical Motion: $a_y = -g$, meaning vertical velocity decreases linearly until reaching zero at the trajectory apex, then accelerates downward toward impact.