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Gravity Velocity Equations for Falling Objects
by Ron Kurtus (revised 5 January 2011)
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 an object falls after a given period of time and the velocity it reaches 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 tools: Metric-English Conversion | Scientific Calculator.
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 Velocity-Distance 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
The following resources provide information on this subject:
Websites
Acceleration due to Gravity Calculations - from Western Washington University
Books
Top-rated
books on Simple Gravity Science
Top-rated
books on Advanced Gravity Physics
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Gravity Velocity Equations for Falling Objects