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Gravity Displacement Equations for Falling Objects
by Ron Kurtus (revised 9 Jun 2011)
The displacement of an object is the change in position from the starting point in a specific direction and can be represented as a vector. It is different from distance, where direction is not indicated.
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 displacement the object falls until it reaches a given velocity or after a given period of time. 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 displacement to reach a given velocity?
- What is the equation for the displacement for a given time?
- What are some examples of these equations?
This lesson will answer those questions.
Useful tool: Metric-English Conversion
Displacement with respect to velocity
The general gravity equation for displacement with respect to velocity is:
y = (v2 − vi2)/2g
(See Derivation of Displacement-Velocity Gravity Equations for details of the derivation.)
Since the initial velocity vi = 0 for a dropped object, the equation reduces to:
y = v2/2g
where
- y is the vertical displacement in meters (m) or feet (ft)
- v is the vertical velocity in m/s or ft/s
- g is the acceleration due to gravity (9.8 m/s2 or 32 ft/s2)
Since the object is moving downward from the starting point, both y and v are positive numbers.
Displacement of a falling object as a function of velocity or time
Displacement with respect to time
The general gravity equation for the displacement with respect to time is:
y = gt2/2 + vit
(See Derivation of Displacement-Time Gravity Equations for details of the derivations.)
Since vi = 0 for a dropped object, the equation reduces to:
y = gt2/2
where t is the time in seconds (s).
Examples
The following examples illustrate applications of the equations.
Given the velocity
If v = 75 ft/s, how far has the object fallen?
Solution
Since v is in ft/s, g = 32 ft/s2. Substitute values in the equation:
y = v2/2g
y = (75 ft/s)*(75 ft/s)/2*(32 ft/s2)
y = (5625 ft2/s2)/(64 ft/s2)
y = 87.89 ft
Given the elapsed time
If t = 4 seconds and g = 9.8 m/s2, find the displacement the object has fallen.
Solution
Substitute values in the equation:
y = gt2/2
y = (9.8 m/s2)*(16 s2)/2
y = 78.4 m
Summary
Displacement is the change in position from the starting point in a specific direction. The following equations allow you to calculate the displacement the object falls until it reaches a given velocity or after a given period of time:
y = v2/2g
y = gt2/2
Have confidence in yourself
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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