Free Fall Distance Calculator — How Far Does an Object Fall?
The Free Fall Distance Calculator computes how far an object drops under gravity alone — with no air resistance — given elapsed time and gravitational acceleration. It applies the kinematic equation h = ½ × g × t², where h is the fall distance in meters, g is the local gravitational acceleration (9.81 m/s² on Earth's surface), and t is the time in seconds. This tool is used in physics education, engineering safety analysis, accident reconstruction, and sports science whenever you need to know how much vertical distance an object (or person) covers during free fall. Because distance grows with the square of time, a fall that lasts twice as long covers four times the distance — a non-obvious fact with critical real-world implications.
An object in free fall drops **h = ½ × g × t²** meters. On Earth (g = 9.81 m/s²): 1 s → 4.9 m, 2 s → 19.6 m, 3 s → 44.1 m (~10 stories), 4 s → 78.5 m (~20 stories), 5 s → 122.6 m. Impact speed is v = g × t.
When to use this calculator
- Estimating the height of a cliff or building by timing a dropped rock until it hits the ground
- Calculating fall distance for workplace fall-protection compliance (OSHA requires arrest within 6 feet / 1.8 m for most general industry)
- Reconstructing accident physics — determining how far a person fell based on witness-reported fall duration
- Designing amusement park free-fall rides (e.g., drop towers) to set ride height for a target seconds-in-air experience
- Comparing free-fall physics on the Moon (g = 1.62 m/s²) vs. Earth for space-mission training simulations
- Verifying skydive altimeter data — checking expected altitude loss during a delayed deployment sequence
Worked Example: 3-Second Fall
- t = 3 s, g = 9.81 m/s²
- h = ½ × 9.81 × 3² = ½ × 9.81 × 9 = 44.145 m
- Impact speed: v = 9.81 × 3 = 29.43 m/s ≈ 65.8 mph
- Equivalent to falling from approximately 10 stories
How it works
3 min readFree Fall Formula
Free fall distance is derived from Newton's second law and constant-acceleration kinematics. Assuming the object starts from rest (initial velocity = 0) and air resistance is negligible:
h = ½ × g × t²
Where:
h = vertical fall distance (meters)
g = gravitational acceleration (m/s²)
Earth surface standard: 9.80665 m/s² (NIST defined)
t = elapsed fall time (seconds)
Derived quantities:
v = g × t (impact velocity, m/s)
v_mph = v × 2.237 (convert to miles per hour)
t = √(2h / g) (time from height)> Example: t = 3 s, g = 9.81 m/s²
> h = 0.5 × 9.81 × 3² = 0.5 × 9.81 × 9 = 44.145 m ≈ 44.1 m
> Impact velocity: v = 9.81 × 3 = 29.43 m/s ≈ 65.8 mph
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Free Fall Distance Reference Table
| Time (s) | Earth (9.81 m/s²) | Moon (1.62 m/s²) | Mars (3.72 m/s²) | Impact Speed (Earth) | Stories* |
|---|---|---|---|---|---|
| 0.5 | 1.23 m | 0.20 m | 0.47 m | 5.5 mph | Ground floor |
| 1.0 | 4.91 m | 0.81 m | 1.86 m | 21.9 mph | 1–2 stories |
| 1.5 | 11.0 m | 1.82 m | 4.19 m | 33.0 mph | 3–4 stories |
| 2.0 | 19.6 m | 3.24 m | 7.44 m | 43.9 mph | 5–6 stories |
| 2.5 | 30.6 m | 5.06 m | 11.6 m | 55.6 mph | 8 stories |
| 3.0 | 44.1 m | 7.29 m | 16.7 m | 65.8 mph | ~10 stories |
| 4.0 | 78.5 m | 12.96 m | 29.8 m | 88.5 mph | ~20 stories |
| 5.0 | 122.6 m | 20.25 m | 46.5 m | 110.4 mph | ~30 stories |
| 6.0 | 176.6 m | 29.16 m | 67.0 m | 134.2 mph | ~45 stories |
| 10.0 | 490.5 m | 81.0 m | 186 m | 220 mph | ~120 stories |
Assuming ~4 m (13 ft) per story average for a commercial building.
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Real-World Cases
Case 1 — Cliff height from a dropped stone
A hiker drops a rock from a ledge and hears the impact after t = 2.5 s.
h = ½ × 9.81 × 2.5² = ½ × 9.81 × 6.25 = 30.6 m (≈ 100 ft)
Impact speed: 9.81 × 2.5 = 24.5 m/s ≈ 54.8 mph — fast enough to be lethal for a person.
Case 2 — OSHA fall arrest distance check
OSHA 29 CFR 1926.502(d) limits free fall to 6 feet (1.83 m) for construction workers.
Time to fall 1.83 m: t = √(2 × 1.83 / 9.81) = √0.373 = 0.61 s
Speed at arrest: 9.81 × 0.61 = 5.99 m/s ≈ 13.4 mph — illustrating why deceleration distance in the lanyard matters enormously.
Case 3 — Drop tower ride design
A theme park targets a 3-second free-fall sensation for riders.
h = ½ × 9.81 × 9 = 44.1 m tower height needed.
The ride structure must be at least ~48–50 m tall to account for deceleration braking distance at the bottom (~5–6 m additional).
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Common Errors
1. Forgetting to square the time — h = ½ × g × t (wrong) vs. h = ½ × g × t² (correct). At t = 3 s the wrong formula gives 14.7 m instead of 44.1 m.
2. Using g = 10 m/s² without noting the approximation — While convenient, it overestimates distance by ~2%. For safety engineering, always use g = 9.80665 m/s² (NIST standard).
3. Ignoring air resistance for long falls — A human body reaches terminal velocity (~56 m/s / 125 mph) after roughly 12–14 seconds. Beyond that, distance is no longer proportional to t².
4. Mixing units — Using time in seconds with g in ft/s² (32.174) and expecting meters. Always keep SI units consistent: g in m/s², t in s → h in meters. To get feet: h_ft = ½ × 32.174 × t².
5. Applying free-fall formula to drag scenarios — Once a parachute deploys or any significant drag force is present, the constant-acceleration model breaks down completely.
Frequently asked questions
What is the standard value of g used in free-fall calculations?
The NIST-defined standard gravitational acceleration is 9.80665 m/s² (exactly), often rounded to 9.81 m/s² for engineering work or 9.8 m/s² for introductory physics. This value applies at sea level at approximately 45° latitude. It can vary from about 9.764 m/s² at the equator to 9.863 m/s² at the poles due to Earth's rotation and shape.
How far does an object fall in the first second of free fall on Earth?
In the first second, h = ½ × 9.81 × 1² = 4.905 m (≈ 16.1 ft). This is often cited as a benchmark: roughly the height of a one-to-two story structure. By second 2, total distance is 19.6 m — four times farther — illustrating the quadratic (not linear) nature of free fall.
Does free fall distance change on the Moon or other planets?
Yes — substantially. On the Moon (g = 1.62 m/s²), an object falls only 0.81 m in 1 second vs. 4.9 m on Earth. On Mars (g = 3.72 m/s²), it falls 1.86 m in 1 second. On Jupiter (g ≈ 24.79 m/s²), the same 1-second fall covers 12.4 m. The calculator lets you input any g value to compare planetary scenarios.
What is terminal velocity and when does the free-fall formula stop being valid?
Terminal velocity is reached when aerodynamic drag equals gravitational force, producing zero net acceleration. For a human body in a belly-to-earth position, terminal velocity is approximately 56 m/s (125 mph), typically reached after 12–14 seconds and roughly 450–500 m of fall. Beyond this point, the h = ½gt² formula significantly overestimates actual fall distance.
How does OSHA use free-fall physics in workplace safety regulations?
OSHA standard 29 CFR 1926.502(d) (construction) limits free fall to 6 feet (1.83 m) before a personal fall arrest system must engage. At that distance, fall time is only ~0.61 s and impact speed reaches ~13.4 mph. The standard also limits total arresting distance (free fall + deceleration) to 3.5 m (11.5 ft) to prevent workers from striking lower surfaces.
How can I use this calculator to estimate a cliff height from a drop time?
Simply time how long an object takes to fall to the ground (e.g., by dropping a stone and listening for impact), then enter that time with g = 9.81 m/s². A 2.0 s drop → 19.6 m; a 2.5 s drop → 30.6 m. For accuracy, repeat the timing 3–5 times and average the results, as human reaction time introduces ~0.2 s of error.
What impact speed corresponds to different fall heights, and how dangerous are they?
Using v = √(2 × g × h): a 3 m fall (about 10 ft, one story) produces v ≈ 7.7 m/s (17 mph); a 10 m fall → v ≈ 14 m/s (31 mph); a 45 m fall (≈10 stories) → v ≈ 29.7 m/s (66 mph). The CDC reports falls as the leading cause of injury-related death in the US, with falls from heights above 6 m (20 ft) carrying a very high fatality risk without protective equipment.
Is the free-fall formula the same whether the object is dropped or thrown horizontally?
Yes — for vertical distance only. An object thrown horizontally still experiences the same downward free-fall acceleration g, so its vertical drop after time t is still h = ½ × g × t², independent of horizontal speed. This is a core principle of projectile motion: horizontal and vertical motions are fully independent when air resistance is neglected.
Why do heavier and lighter objects fall at the same rate in free fall?
Galileo's equivalence principle states that gravitational acceleration is independent of mass. The gravitational force on an object is F = m × g, but Newton's second law gives a = F/m = g, so mass cancels out. In vacuum, a feather and a bowling ball fall identically — as famously demonstrated on the Moon by Apollo 15 astronaut David Scott in 1971.
How do I calculate free fall distance in feet instead of meters?
Use g = 32.174 ft/s² instead of 9.81 m/s² in the calculator. Then h_ft = ½ × 32.174 × t². For example, at t = 3 s: h = ½ × 32.174 × 9 = 144.8 ft (≈ 44.1 m × 3.281). Alternatively, multiply any meter result by 3.281 to convert to feet.