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Gravity Distance Equations for Objects Projected Downward
by Ron Kurtus (revised 9 August 2010)
When you throw or project an object downward, it has an initial velocity greater than zero (vi > 0). If you know the initial velocity, there are simple derived equations that allow you to calculate the distance traveled from the starting point when the object reaches a given velocity or when it reaches a given elapsed time. Some examples illustrate these equations.
Note: You normally do not need to memorize these equations, but you should know where to find them in order to solve equations.
Questions you may have include:
- How do you find the distance for a given velocity?
- How do you find the distance for a given time?
- What are some examples of these equations?
This lesson will answer those questions. There is a mini-quiz near the end of the lesson.
Useful tools: Metric-English Conversion | Scientific Calculator.
Distance for a given velocity
If you throw an object downward at some initial velocity, the equation for the distance traveled to reach a given velocity is:
y = (v2 − vi2)/2g
where
- y is the vertical distance from the starting point in meters (m) or feet (ft)
- v is the vertical velocity in meters/second (m/s) or feet/second (ft/s)
- vi is the initial vertical velocity in m/s or ft/s
- g is the acceleration due to gravity (9.8 m/s2 or 32 ft/s2)
(See Derivation of Velocity-Distance Gravity Equations for details of the derivations.)
Distance of an object thrown downward as a function of
initial velocity and velocity or time
Distance for a given time
If you throw an object downward, the equation for the distance traveled within a given time is:
y = gt2/2 + vit
where t is the time the object has fallen in seconds (s).
(See Derivation of Distance-Time Gravity Equations for details of the derivations.)
Examples
The following examples illustrate applications of the equations.
For a given velocity
If you throw an object downward at 10 m/s, find the minimum height you must throw it from so that it reaches 50 m/s.
Solution
You are given that vi = +10 m/s and v = 50 m/s. Since vi and v are in m/s,
g = 9.8 m/s2. The equation to use is:
y = (v2 − vi2)/2g
Substitute values in the equation:
y = [(50 m/s)2 − (10 m/s)2]/2*(9.8 m/s2)
y = [(2500 m2/s2) − (100 m2/s2)]/(19.6 m/s2)
y = (2400 m2/s2)/(19.6 m/s2)
y = 122.4 m
For a given time
If you throw an object downward at 30 ft/s and it travels for 4 seconds, find the distance traveled.
Solution
You are given that vi = 30 ft/s and t = 4 s. Since vi is in ft/s, g = 32 ft/s2. The equation to use is:
y = gt2/2 + vit
Substitute values in the equation:
y = [(32 ft/s2)*(4 s)2]/2 + (30 ft/s)*(4 s)
y = (32 ft/s2)*(16 s2)/2 + 120 ft
y = (512 ft)/2 + 120 ft
y = 256 ft + 120 ft
y = 376 ft
Summary
You can calculate the distance traveled from the starting point when an object that is projected downward reaches a given velocity or when it reaches a given elapsed time from the equations:
y = (v2 − vi2)/2g
y = gt2/2 + vit
See the Side Menu for more Gravity and Gravitation topics
Check your numbers
Resources
The following resources provide information on this subject:
Websites
Acceleration due to Gravity Calculations - from Western Washington University
Gravity and Gravitation Resources
Books
Top-rated
books on Simple Gravity Science
Top-rated
books on Advanced Gravity Physics
Mini-quiz to check your understanding
If you got all three correct, you are on your way to becoming a Champion in Physics. If you had problems, you had better look over the material again.
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Also see Answers to Readers' Questions.
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