# Velocity Equations for Falling Objects

by Ron Kurtus (revised 31 March 2017)

When you drop an object from some height above the ground, it has an initial velocity of zero. Simple equations allow you to calculate the velocity a falling object reaches after a given period of time and its velocity at a given displacement. The equations assume that air resistance is negligible.

Examples demonstrate applications of the equations.

Questions you may have include:

• What is the equation for the velocity for a given time?
• What is the equation for the velocity to reach a given displacement?
• What are some examples of these equations?

This lesson will answer those questions. Useful tool: Units Conversion

## Velocity with respect to time

The general gravity equation for velocity with respect to time is:

v = gt + vi

(See Derivation of Velocity-Time Gravity Equations for details of the derivation.)

Since the initial velocity vi = 0 for an object that is simply falling, the equation reduces to:

v = gt

where

• v is the vertical velocity of the object in meters/second (m/s) or feet/second (ft/s)
• g is the acceleration due to gravity (9.8 m/s2 or 32 ft/s2)
• t is the time in seconds (s) that the object has fallen Velocity of a falling object as a function of time or displacement

## Velocity with respect to displacement

The general gravity equation for velocity with respect to displacement is:

v = ±√(2gy + vi2)

where

• ± means plus or minus
• √(2gy + vi2) is the square root of the quantity (2gy + vi2)
• y is the vertical displacement in meters (m) or feet (ft)

(See Derivation of Displacement-Velocity Gravity Equations for details of the derivations.)

Since vi = 0, y is positive because it is below the starting point. Also, v is downward and positive. Only the + term of ± applies.

Thus, the equation for the velocity of a falling object after it has traveled a certain displacement is:

v = √(2gy)

## Examples

The following examples illustrate applications of the equations.

### For a given time

What will be the velocity of an object after it falls for 3 seconds?

#### Solution

Substitute in the equation:

v = gt

If you use g = 9.8 m/s2, v = (9.8 m/s2)*(3 s) = 29.4 m/s.

If you use g = 32 ft/s2, v = (32 ft/s2)*(3 s) = 96 ft/s.

### For a given displacement

What is the velocity of an object after it has fallen 100 feet?

#### Solution

Since y is in feet, g = 32 ft/s2. Substitute in the equation:

v = √(2gy)

v = √[2*(32 ft/s2)*(100 ft)]

v = √(6400 ft2/s2)

v = 80 ft/s

## Summary

There are simple equations for falling objects that allow you to calculate the velocity the object reaches for a given displacement or time. The equations are:

v = gt

v = √(2gy)

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## Resources and references

Ron Kurtus' Credentials

### Websites

Gravity Resources

Falling Bodies - Physics Hypertextbook

Equations for a falling body - Wikipedia

Gravity Calculations - Earth - Calculator

Kinematic Equations and Free Fall - Physics Classroom

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