Free Fall Calculator: Velocity, Distance & Time

Free Fall Calculator: Velocity, Distance, and Time

A free fall calculator finds the speed of an object falling under gravity, how far it has moved, and how high it still is above the ground. Enter a starting height, a starting velocity, and an elapsed time, and it returns the final velocity, the downward displacement, and the final height.

What Is Free Fall?

Free fall is motion caused by gravity alone. Air resistance is ignored. Near Earth's surface, every falling object speeds up at the same rate, no matter its mass. A coin and a hammer dropped together hit the ground at the same instant, as long as air drag is small enough to ignore.

Galileo Galilei showed this experimentally in the early 1600s, overturning the older idea that heavier objects fall faster. In 1971, Apollo 15 astronaut David Scott dropped a hammer and a feather on the Moon, where there is no air to slow the feather. Both landed together, on camera.

The rate of that acceleration is called standard gravity, written g. Its defined value is exactly 9.80665 meters per second squared (m/s²), set by international agreement in 1901. In imperial units the same acceleration is 32.17405 feet per second squared (ft/s²). That figure is 9.80665 divided by 0.3048, the exact number of meters in one foot, rounded to five decimal places. Many textbook examples round the metric value to 9.81 m/s² for simpler arithmetic.

Free Fall Formula

Three equations cover every free fall problem. Downward is treated as the positive direction, so a positive velocity points toward the ground and a negative one points up.

Velocity formula

$v = v_0 + gt$

• $v$ = final velocity
• $v_0$ = initial velocity (0 if dropped from rest; negative if thrown upward)
• $g$ = gravitational acceleration, 9.80665 m/s² or 32.17405 feet/s² in this calculator
• $t$ = time elapsed, in seconds

Distance formula

$d = v_0t + \frac{1}{2}gt^2$

Here $d$ is the displacement: how far the object has moved downward. For an object dropped from rest ($v_0 = 0$), the equation simplifies to $d = \frac{1}{2}gt^2$. If the object was thrown upward and is still above its release point, $d$ comes out negative.

Velocity-distance formula

When time is unknown, this equation connects velocity and distance directly:

$v^2 = v_0^2 + 2gd$

For an object dropped from rest, this simplifies to $v = \sqrt{2gd}$. It is the formula used to find impact speed from a known drop height, or drop height from a known impact speed.

How to Calculate Free Fall With This Calculator

The calculator takes four settings and returns three results.

1. Choose a unit system. Metric uses meters and meters per second. Imperial uses feet and feet per second.
2. Enter the initial height, the height the object starts at above the ground. It cannot be negative.
3. Enter the initial velocity. Use 0 for an object dropped from rest, a positive number for an object thrown downward, or a negative number for an object thrown upward.
4. Enter the time elapsed since the object was released. It must be greater than zero.

The results are the final velocity, the displacement, and the final height (the initial height minus the displacement). Time is an input here, not a result: the calculator does not solve backward from a target speed or distance.

If the entered time is long enough that the object would already have hit the ground, the calculator reports that instead of a negative final height.

Example: The Default Setup

An object is dropped from rest at a height of 100 meters. Where is it after 3 seconds?

Velocity after 3 seconds:

$v = 0 + 9.80665 \times 3 = 29.41995 \approx 29.42 \text{ m/s}$

Displacement in those 3 seconds:

$d = \frac{1}{2} \times 9.80665 \times 3^2 = 44.129925 \approx 44.13 \text{ m}$

Final height:

$100 - 44.129925 = 55.870075 \approx 55.87 \text{ m}$

So the object is falling at about 29.42 m/s, has dropped about 44.13 m, and is still about 55.87 m above the ground.

Example: Time and Velocity From a Known Height

A wrench is dropped from a scaffold 20 meters high. How long does it take to hit the ground, and how fast is it moving on impact?

Using $d = \frac{1}{2}gt^2$ with $d = 20$ m and $g = 9.81$ m/s²:

$t = \sqrt{\frac{2 \times 20}{9.81}} \approx 2.02 \text{ seconds}$

$v = gt = 9.81 \times 2.02 \approx 19.8 \text{ m/s (about 71 km/h)}$

Example: Height From Impact Velocity

An object hits the ground at 25 m/s after being dropped from rest. From what height did it fall?

Using $v^2 = 2gd$:

$d = \frac{v^2}{2g} = \frac{25^2}{2 \times 9.81} = \frac{625}{19.62} \approx 31.9 \text{ meters}$

That is roughly the height of a ten-story building.

How Gravity Changes

Standard gravity, 9.80665 m/s², is an average reference value. Actual gravity varies slightly across Earth's surface. It is weaker at the equator (about 9.78 m/s²) than at the poles (about 9.83 m/s²), because Earth bulges outward at the equator and spins fastest there. Gravity also weakens slightly with altitude: at 10,000 meters, it drops to about 9.78 m/s².

Other worlds have very different gravity, so the same formulas give very different results:

The Moon's gravity is about one sixth of Earth's, so an object dropped there covers about one sixth as much distance in the same time.

Free Fall and Air Resistance

These formulas assume a vacuum, with no air pushing back on the object. Real air resistance grows with the square of speed, so it matters more at higher velocities. It has little effect on a dense, compact object like a wrench or a stone falling a short distance. It has a large effect on light or wide objects, such as a feather, a sheet of paper, or a parachute.

When air resistance grows large enough to match the pull of gravity, the object stops accelerating and falls at a constant speed, called terminal velocity. A skydiver in a belly-down position typically reaches a terminal velocity near 53 meters per second, about 190 km/h or 120 mph, before opening a parachute.

Frequently asked questions

What is free fall in physics?

Free fall is the motion of an object acted on by gravity alone, with air resistance ignored. All falling objects near Earth's surface accelerate downward at the same rate, about 9.81 m/s², regardless of their mass.

Why do all objects fall at the same rate?

Gravity pulls harder on a heavier object, but a heavier object also resists a change in motion more (it has more inertia). These two effects cancel exactly, so mass does not affect the rate of acceleration in free fall.

What is the formula for free fall velocity?

$v = v_0 + gt$, where $v_0$ is the starting velocity, $g$ is gravitational acceleration, and $t$ is time. For an object dropped from rest, this is simply $v = gt$.

What is the formula for free fall distance?

$d = v_0t + \frac{1}{2}gt^2$. For an object dropped from rest, this simplifies to $d = \frac{1}{2}gt^2$.

Why is the displacement sometimes negative?

Downward counts as positive in this calculator. A negative displacement means the object is above the point it was released from, which happens when it was thrown upward and has not fallen back yet. Throw a ball up at 10 m/s and check it after 1 second: $d = -10 \times 1 + \frac{1}{2} \times 9.80665 \times 1^2 = -5.10$ m, so it is 5.10 m higher than where it started.

How long does it take an object to fall a given height?

Solving the distance formula for time gives $t = \sqrt{2d/g}$, for an object dropped from rest. A 20-meter drop takes about 2.02 seconds.

Why does this calculator use 9.80665 m/s² instead of 9.8?

9.80665 m/s² is the exact internationally defined value of standard gravity, adopted in 1901. Rounding it to 9.8 or 9.81 is common in textbooks and is accurate enough for most purposes. The calculator uses the defined value so that metric and imperial results describe the same physical fall, which is not true of the rounded pair 9.8 m/s² and 32.2 feet/s².

Does this calculator account for air resistance?

No. It assumes idealized free fall in a vacuum. This gives accurate results for dense, compact objects falling moderate distances, but overestimates speed and distance for light or wide objects, or for very long falls where air resistance becomes significant.