Calculate horizontal projectile range (R = v0·√(2h/g)), time of flight (t = √(2h/g)), vertical impact velocity (vy = -√(2gh)), landing speed (v = √(v0² + 2gh)), and trajectory angle with the free Horizontal Projectile Motion Calculator.
16.40 ft • Flight Time: 1.43 s • Impact Speed: 14.44 m/s (52.0 km/h) • Angle: -75.9°
√(2h0 / g)
52.0 km/h (32.3 mph)
-√(2gh0)
Below horizontal
1/2 · m · v0²
Ek_init + m·g·h0
In classical Newtonian kinematics, horizontal projectile motion represents a special case of two-dimensional motion where an object is launched from an initial elevation \(h_0\) with a purely horizontal initial velocity (\(v_{0x} = v_0\)) and zero initial vertical velocity (\(v_{0y} = 0\)). As first demonstrated by Galileo Galilei, the horizontal and vertical motions are strictly independent (orthogonal):
With no horizontal forces acting (\(a_x = 0\)), horizontal velocity remains strictly invariant throughout flight: \(v_x(t) = v_0 \implies x(t) = v_0 t\).
Under uniform gravitational acceleration (\(a_y = -g\)), the object accelerates downward identically to a dropped body: \(y(t) = h_0 - \frac{1}{2} g t^2\).
Combining both components yields a semi-parabolic trajectory curve: \(y(x) = h_0 - \frac{g}{2 v_0^2} x^2\).
Summary of foundational equations across physics, forensics, and aerospace engineering:
| Physical Metric | Mathematical Formula | SI Units & Description |
|---|---|---|
| Total Time of Flight (\(t_{\text{flight}}\)) | $$t_{\text{flight}} = \sqrt{\frac{2 h_0}{g}}$$ | \(\text{seconds (s)}\) |
| Horizontal Range (\(R = x_{\text{max}}\)) | $$R = v_0 t_{\text{flight}} = v_0 \sqrt{\frac{2 h_0}{g}}$$ | \(\text{meters (m)}\) |
| Vertical Landing Velocity (\(v_y\)) | $$v_y = -g t_{\text{flight}} = -\sqrt{2 g h_0}$$ | \(\text{m/s}\) (Downward) |
| Total Resultant Impact Speed (\(v_{\text{impact}}\)) | $$v_{\text{impact}} = \sqrt{v_0^2 + v_y^2} = \sqrt{v_0^2 + 2 g h_0}$$ | \(\text{m/s}\) |
| Impact Landing Angle (\(\theta_{\text{impact}}\)) | $$\theta_{\text{impact}} = \arctan\left(\frac{|v_y|}{v_0}\right) = \arctan\left(\frac{\sqrt{2 g h_0}}{v_0}\right)$$ | \(\text{degrees (^\circ)}\) |
| Parabolic Trajectory Function | $$y(x) = h_0 - \frac{g}{2 v_0^2} x^2$$ | Path in vertical plane |
Solve from initial launch velocity & height, target range & height, or query instantaneous state at any intermediate time \(t\).
Input horizontal launch speed \(v_0\) (in \(\text{m/s}\), \(\text{km/h}\), \(\text{mph}\), or \(\text{ft/s}\)) and initial elevation \(h_0\).
Choose Earth (\(9.807\text{ m/s}^2\)), Moon (\(1.62\text{ m/s}^2\)), Mars (\(3.71\text{ m/s}^2\)), Jupiter, or custom gravity.
Inspect total horizontal range, time of flight, resultant landing speed, impact angle, and live KaTeX mathematical proofs.
One of the most famous counterintuitive demonstrations in physics:
If a rifle fires a bullet horizontally at \(900\text{ m/s}\) from a height of \(1.5\text{ m}\) while a second bullet is dropped simultaneously from rest at the same height, both bullets strike the ground at the exact same instant (\(t = \sqrt{2 \times 1.5 / 9.807} \approx 0.553\text{ seconds}\)).
Because gravity exerts force exclusively in the vertical dimension (\(a_y = -g\)), horizontal velocity has zero component along the vertical axis. The fired bullet travels \(R = 900 \times 0.553 \approx 498\text{ meters}\) horizontally before landing, while the dropped bullet travels \(0\text{ meters}\), yet their vertical descents are identical.
How accident reconstruction experts determine vehicle speed prior to vaulting off bridges or drop-offs:
When a vehicle launches horizontally off a cliff of height \(h_0\) and lands at horizontal distance \(R\), investigators reconstruct the take-off velocity by eliminating flight time:
Solve for horizontal range (\(R\)), required launch speed (\(v_0\)), required drop height (\(h_0\)), or intermediate points.
Computes resultant landing speed (\(v_{\text{impact}}\)), vertical velocity (\(v_y\)), and impact angle below the horizontal.
Evaluate horizontal trajectories on Earth (\(9.807\text{ m/s}^2\)), Moon (\(1.62\text{ m/s}^2\)), Mars (\(3.71\text{ m/s}^2\)), or custom worlds.
Tracks initial and impact kinetic energy: \(E_{k,\text{impact}} = E_{k,\text{init}} + m g h_0\).
Renders clear algebraic proofs with live numeric substitutions for physics coursework.
Runs instantly on any smartphone, tablet, or desktop with zero server lag and total calculation privacy.
How transport aircraft calculate the release point to land cargo packages on target:
A cargo plane flying level at altitude \(h_0 = 500\text{ m}\) and cruise speed \(v_0 = 80\text{ m/s}\) (\(288\text{ km/h}\)) must release a package well before flying over the target drop zone:
Calculating vertical trajectory drop over flat shooting ranges:
When a rifle barrel is leveled horizontally at distance \(x\), the vertical gravitational drop \(\Delta y\) scales quadratically with target distance:
For a high-velocity rifle bullet (\(v_0 = 900\text{ m/s}\)), the drop at \(100\text{ m}\) is only \(6.05\text{ cm}\), while at \(300\text{ m}\) it increases ninefold to \(54.5\text{ cm}\).
How planetary surface gravity scales horizontal projectile range:
Standard terrestrial baseline. A \(10\text{ m}\) drop gives \(t = 1.43\text{ s}\) and range \(R = 1.43 v_0\).
Lower gravity extends flight time by \(2.46\times\), yielding \(t = 3.51\text{ s}\) and range \(R = 3.51 v_0\).
Intermediate gravity extends flight time by \(1.63\times\), yielding \(t = 2.32\text{ s}\) and range \(R = 2.32 v_0\).
Why real-world projectiles fall short of theoretical vacuum trajectories:
In real atmospheres, air resistance exerts a continuous retarding drag force opposing the instantaneous velocity vector: \(F_d = \frac{1}{2} \rho C_d A v^2\). This steadily diminishes horizontal velocity \(v_x(t) < v_0\) and establishes a vertical terminal velocity limit \(v_{\text{term}} = \sqrt{\frac{2 m g}{\rho C_d A}}\). Dense, aerodynamic projectiles closely match vacuum equations, whereas lightweight objects experience significant range reduction.
Calculating flight distance when launching horizontally over an inclined landing hill:
When a ski jumper leaves the takeoff table horizontally at velocity \(v_0\) above an inclined landing hill sloping downward at angle \(\theta\), the landing point is the geometric intersection of the parabolic trajectory with the slope line \(y(x) = -x \tan\theta\):
Olympic jump hills are curved to match the jumper's parabolic descent path, ensuring a parallel landing angle that minimizes vertical impact shock.
Calculating horizontal range for fluid streams exiting a punctured water tank:
For an orifice at depth \(h\) below the surface of a tank with total height \(H\): \(R = 2\sqrt{h(H - h)}\).
Differentiating with respect to depth shows that maximum horizontal jet range occurs when the hole is placed exactly at half height: \(h = H / 2 \implies R_{\text{max}} = H\).
Determining horizontal spread swaths in agricultural and industrial machinery:
In spinning disc broadcast seeders and wheelabrator shot peening machines, particles depart the spinning disc edge at tangential velocity \(v_0 = \omega r\). From mounting height \(h_0\), the horizontal throw radius is:
Comprehensive answers to common questions about horizontal projectile formulas, flight time, range, impact velocity, and trajectory physics.
