42.78
m/s
154.02
km/h
95.7
mph
784.8
N
42.78
m/s
154.02
km/h
95.7
mph
784.8
N
The Terminal Velocity Calculator determines the maximum speed a falling object reaches when aerodynamic drag equals gravitational force. At terminal velocity, net force is zero and the object stops accelerating, descending at a constant speed determined by its mass, shape, and the surrounding fluid properties.
Terminal velocity is given by the equation: $$v_t = \sqrt{\frac{2mg}{\rho C_d A}}$$ where $$m$$ is mass, $$g$$ is gravitational acceleration, $$\rho$$ is fluid density, $$C_d$$ is the drag coefficient, and $$A$$ is the reference cross-sectional area perpendicular to the flow.
The drag coefficient $$C_d$$ depends on the object's shape and the Reynolds number of the flow. A sphere has $$C_d \approx 0.47$$, a flat plate perpendicular to flow has $$C_d \approx 1.17$$, and a streamlined body can have $$C_d \approx 0.04$$. A skydiver in belly-down position has $$C_d \approx 1.0$$ with cross-sectional area around 0.7 m², while in a head-down dive the area drops to about 0.15 m², dramatically increasing terminal velocity.
Air density varies significantly with altitude: at sea level $$\rho \approx 1.225\,\text{kg/m}^3$$, but at 4,000 m altitude it drops to about 0.82 kg/m³. This means terminal velocity increases at higher altitudes—Felix Baumgartner reached supersonic speeds during his 2012 stratospheric jump partly because the near-vacuum upper atmosphere provided negligible drag initially.
The concept extends beyond skydivers to raindrops (terminal velocity 1–9 m/s depending on size), hailstones, parachute design, seed dispersal, volcanic ash fallout, and sedimentation in fluids. Engineers use terminal velocity calculations in designing aerodynamic vehicles, ballistic trajectories, and industrial settling tanks for particle separation.
This calculator assumes the standard drag equation valid for high Reynolds number flow. For very small particles (dust, fog droplets) where Reynolds numbers are below 1, Stokes' law applies instead: $$v_t = 2r^2 g(\rho_p - \rho_f) / (9\mu)$$, which is proportional to radius squared rather than mass.
Enter the object's mass, drag coefficient, cross-sectional area, and air density. The calculator applies $$v_t = \sqrt{2mg / (\rho C_d A)}$$ to find terminal velocity. At this speed, drag force exactly equals gravitational force ($$F_d = mg$$), so the results also show the drag force magnitude and velocity in km/h and mph for convenience.
Higher mass or lower drag area yields faster terminal velocity. The drag force at terminal velocity equals the object's weight (mg). Actual approach to terminal velocity is exponential—an object reaches ~63% of terminal velocity after one characteristic time constant $$\tau = m/(\rho C_d A v_t / 2)$$, and 95% after about three time constants.
Inputs
Results
An 80 kg skydiver in belly-down position reaches ~43 m/s (154 km/h) terminal velocity.
Inputs
Results
A golf ball (45.9 g, Cd=0.47, A=14.32 cm²) has terminal velocity ~33 m/s.
Terminal velocity is the constant maximum speed reached by a falling object when the drag force equals the gravitational force, resulting in zero net acceleration.
Yes, in air. A heavier object has higher terminal velocity because gravitational force increases linearly with mass while drag depends on cross-sectional area and shape, not mass directly.
About 55 m/s (200 km/h) belly-down and up to 90 m/s (320 km/h) in a head-down dive. With a parachute deployed, it drops to about 5 m/s.
Higher altitude means lower air density, which increases terminal velocity. At 10,000 m altitude, air density is about one-third of sea level, so terminal velocity increases by roughly 70%.
A dimensionless number representing aerodynamic resistance. Sphere: 0.47, flat plate: 1.17, streamlined body: 0.04. It depends on shape, surface roughness, and Reynolds number.
Yes, in thin atmosphere. Felix Baumgartner reached Mach 1.25 during his 2012 stratospheric jump because air density at 39 km altitude is less than 1% of sea-level density.
Roboculator Team
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