Terminal Velocity Calculator
Free Terminal velocity Calculator for mechanics. Enter variables to compute results with formulas and detailed steps. Get results you can export or share.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator
Terminal Velocity Calculator
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Formula: Vt = sqrt(2mg / (rho x Cd x A))
Worked example — Terminal Velocity: 51.14 m/s (184 km/h / 114 mph) | Time to 95%: ~15.6s | Distance: ~540m
Formula
Vt = sqrt(2mg / (rho x Cd x A))
Terminal velocity equals the square root of (2 times mass times gravitational acceleration) divided by (air density times drag coefficient times cross-sectional area). This is derived by setting gravitational force (mg) equal to drag force (0.5 x rho x Cd x A x V^2) and solving for velocity.
Worked Examples
Example 1: Skydiver in Belly-Down Position
Problem:An 80 kg skydiver falls in belly-down position with Cd = 0.7 and cross-sectional area = 0.7 m2 at sea level (air density = 1.225 kg/m3).
Solution:Vt = sqrt(2 x 80 x 9.81 / (1.225 x 0.7 x 0.7)) Vt = sqrt(1569.6 / 0.60025) Vt = sqrt(2615.1) Vt = 51.14 m/s = 184 km/h = 114 mph Weight = 80 x 9.81 = 784.8 N Drag force at terminal velocity = 784.8 N Time to 95% = (51.14/9.81) x 3.0 = 15.6 seconds Distance to 95% = approx 540 m
Result:Terminal Velocity: 51.14 m/s (184 km/h / 114 mph) | Time to 95%: ~15.6s | Distance: ~540m
Example 2: Golf Ball Free Fall
Problem:A golf ball (mass = 0.046 kg, diameter = 42.67mm, Cd = 0.25) falls from a tall building. Calculate terminal velocity at sea level.
Solution:Area = pi x (0.02134)^2 = 0.001431 m2 Vt = sqrt(2 x 0.046 x 9.81 / (1.225 x 0.25 x 0.001431)) Vt = sqrt(0.9025 / 0.000438) Vt = sqrt(2060.5) Vt = 45.39 m/s = 163 km/h = 101 mph Weight = 0.046 x 9.81 = 0.451 N Mach = 45.39 / 343 = 0.132
Result:Terminal Velocity: 45.39 m/s (163 km/h / 101 mph) | Mach: 0.132 | Subsonic
Frequently Asked Questions
What is terminal velocity and how is it reached?
Terminal velocity is the maximum speed an object reaches when falling through a fluid (like air) when the drag force equals the gravitational force, resulting in zero net acceleration. When you first release an object, gravity accelerates it downward at 9.8 m/s per second. As speed increases, air resistance (drag force) grows proportionally to the square of velocity. At some point, drag exactly balances gravity and the object stops accelerating, falling at a constant speed. This speed is terminal velocity. A skydiver in a belly-down position reaches about 120 mph (195 km/h) after approximately 12 seconds and 450 meters of free fall. The formula is Vt = sqrt(2mg / (rho x Cd x A)), derived from setting the gravitational force equal to the drag force equation.
What factors affect terminal velocity?
Four primary factors determine terminal velocity. Mass (or weight) is directly proportional to terminal velocity: heavier objects fall faster because they need more drag force to balance gravity. The drag coefficient depends on the object shape: streamlined shapes like teardrops (Cd = 0.04) have much lower drag than flat plates (Cd = 1.28). Cross-sectional area is the projected area perpendicular to the direction of motion. A skydiver belly-down presents about 0.7 square meters, while head-down presents only 0.3 square meters, dramatically increasing speed. Air density decreases with altitude: at 30,000 feet, air density is roughly 40% of sea level, so terminal velocity is much higher at altitude. This is why Felix Baumgartner exceeded the speed of sound during his stratospheric jump, where air density was extremely low.
What is the terminal velocity of a human body?
A human skydiver has different terminal velocities depending on body position and equipment. In the standard belly-to-earth spread position, terminal velocity is approximately 120 mph (195 km/h or 54 m/s) for an average 80 kg person with a drag coefficient of about 0.7 and cross-sectional area of 0.7 square meters. In a head-down dive position, the reduced area (about 0.3 square meters) and lower drag coefficient (about 0.4) increase terminal velocity to approximately 200 mph (320 km/h). Competitive speed skydivers wearing tight suits in a head-down position can exceed 300 mph (480 km/h). With a deployed parachute (area approximately 30 square meters, Cd approximately 1.5), terminal velocity drops to about 12-18 mph (5-8 m/s), allowing safe landing. Body weight significantly affects speed, with heavier skydivers falling faster.
How does the drag coefficient vary for different shapes?
The drag coefficient (Cd) is a dimensionless number that quantifies the drag of an object relative to a theoretical flat plate of the same area. A sphere has Cd approximately 0.47, while a streamlined airfoil shape has Cd approximately 0.04, showing the massive impact of shape optimization. Common values include: flat plate perpendicular to flow is 1.28, a cube is 1.05, a cylinder across the flow is 1.2, a cone pointing forward is 0.5, a bullet shape is 0.295, and a teardrop is 0.04. For practical objects: a skydiver belly-down is about 0.7-1.0, a car is 0.25-0.45, a bicycle rider is 0.9, and a golf ball (with dimples) is 0.25 versus a smooth ball at 0.47. The dimples on a golf ball actually reduce drag by triggering turbulent boundary layer flow that delays separation, allowing the ball to travel farther.
How does altitude and air density affect terminal velocity?
Air density decreases exponentially with altitude, and since terminal velocity is inversely proportional to the square root of density, objects fall faster at higher altitudes. At sea level, air density is 1.225 kg/m3. At 5,000 meters (16,400 feet), it drops to about 0.736 kg/m3, increasing terminal velocity by roughly 29%. At 10,000 meters (32,800 feet, typical cruising altitude), density is about 0.414 kg/m3, increasing terminal velocity by about 72%. At 39,000 meters (Felix Baumgartner jump altitude), air density is only about 0.004 kg/m3, making terminal velocity roughly 17 times higher than at sea level. This is how Baumgartner reached 833 mph (Mach 1.25) during his fall. As he descended into denser air, drag increased and he naturally decelerated without any action needed. Temperature also affects density: hot air is less dense, slightly increasing terminal velocity on warm days.
What is the physics behind reaching terminal velocity?
The approach to terminal velocity follows a hyperbolic tangent function. The exact equation for velocity as a function of time is v(t) = Vt x tanh(g x t / Vt), where Vt is terminal velocity, g is gravitational acceleration, and t is time. Initially (small t), tanh approximates its argument, so v approximately equals g x t, showing free-fall acceleration as if no drag existed. As t grows large, tanh approaches 1, and velocity asymptotically approaches Vt. The object reaches 50% of terminal velocity in about 0.55 x Vt/g seconds, 90% in about 1.47 x Vt/g seconds, and 95% in about 1.83 x Vt/g seconds. Theoretically, terminal velocity is never exactly reached, only approached asymptotically. For a skydiver with Vt = 54 m/s, reaching 95% (51 m/s) takes about 10 seconds and roughly 350 meters of fall distance. The distance equation involves the natural log of cosh, making analytical solutions possible.
How do parachutes work to reduce terminal velocity?
Parachutes reduce terminal velocity by dramatically increasing both the cross-sectional area and drag coefficient. A typical sport parachute has a canopy area of 20-30 square meters (compared to a skydiver body of 0.7 square meters) and a drag coefficient of 1.3-1.5 (compared to 0.7 for a body). Plugging these into the terminal velocity formula, the speed drops from about 54 m/s (120 mph) to about 5-7 m/s (11-16 mph), which is a survivable landing speed. Round parachutes (military style) have higher drag but no forward glide capability. Ram-air rectangular parachutes (sport skydiving) generate lift as well as drag, allowing forward flight at 20-40 mph with controlled descent rates. Drogue chutes (small stabilizing parachutes) are used to reduce speed partially from very high velocities. Parachute deployment timing is critical because opening at too high a speed can damage the canopy or cause injury from the sudden deceleration forces.
Can objects exceed terminal velocity?
Yes, objects can exceed terminal velocity in several scenarios. If an object enters a denser medium after accelerating in a less dense one (like a meteoroid entering the atmosphere at cosmic velocity of 11-72 km/s, far exceeding atmospheric terminal velocity), it will be traveling faster than terminal velocity and will decelerate. An object thrown or propelled downward can initially exceed terminal velocity until drag slows it to the terminal speed. Changing body position while falling can temporarily exceed the new terminal velocity. For example, a skydiver at 120 mph who transitions from belly-down to a tighter tuck position may briefly exceed the tuck position terminal velocity of 200+ mph. Objects falling through decreasing density (like falling from high altitude toward space with decreasing atmospheric density, which does not apply on Earth) could continuously accelerate. In everyday scenarios on Earth, terminal velocity acts as a natural speed limit for freely falling objects.
How is terminal velocity used in engineering applications?
Terminal velocity concepts are essential across many engineering fields. In aerospace, understanding drag and terminal velocity helps design reentry vehicles that manage deceleration forces. The Mars Curiosity rover used a supersonic parachute sized for the thin Martian atmosphere where terminal velocity is much higher than on Earth. In environmental engineering, terminal velocity of particles determines settling rates in water treatment plants and air pollution control. Stokes Law (for low Reynolds number) calculates terminal velocity of small particles like dust, fog droplets, and sediment. In sports engineering, golf ball dimple design, cycling helmets, and swimsuit textures are optimized using drag coefficient analysis. In industrial processes, spray drying, pneumatic conveying, and fluidized beds all depend on particle terminal velocity. Wind turbine blade design balances lift and drag forces. Even rain drop terminal velocity (about 9 m/s for large drops) matters for erosion studies and aircraft design.
What is the difference between terminal velocity in air versus water?
The fundamental physics is identical, but water is roughly 800 times denser than air (1,000 kg/m3 versus 1.225 kg/m3), dramatically reducing terminal velocity. A steel ball bearing that falls at 40 m/s in air reaches only about 1.4 m/s in water. Additionally, water has much higher viscosity than air, and for small particles (low Reynolds number), viscous forces dominate over inertial forces, making the drag relationship linear rather than quadratic. This is described by Stokes Law: Vt = (2r^2 x (rho_particle - rho_fluid) x g) / (9 x mu), where r is particle radius, rho values are densities, and mu is dynamic viscosity. In water, buoyancy also plays a significant role since the displaced fluid weight partially counteracts gravity. Objects less dense than water (like wood) never reach terminal velocity downward; instead, they reach terminal velocity upward as they float. Submarines and underwater vehicles operate at speeds far below the terminal velocity of their hull shape.
References
Background & Theory
Terminal Velocity of Common Falling Objects
Terminal velocity is reached when drag force (which grows with the square of speed) exactly balances gravitational force, so the object stops accelerating and falls at a constant rate. Because drag depends on cross-sectional area and shape as much as on mass, two objects with identical weight can have very different terminal velocities depending on how they're oriented.
| Falling object | Approx. terminal velocity |
|---|---|
| Raindrop | ~9 m/s (32 km/h) |
| Tennis ball | ~31 m/s (110 km/h) |
| Baseball | ~42 m/s (150 km/h) |
| Skydiver, belly-to-earth | ~53-55 m/s (190-200 km/h) |
| Skydiver, head-down dive | ~90 m/s (320 km/h) |
This is exactly why a skydiver can control descent speed mid-air without any equipment: flattening out into a belly-down "arch" position roughly doubles the cross-sectional area facing the airflow compared to a head-down dive, which alone accounts for most of the difference between the two terminal velocities above.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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