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Gravity Velocity Equations for Objects Projected Upward
by Ron Kurtus (26 August 2009)
If you know the initial velocity that an object is thrown upward, you can calculate its velocity at any time. You can also determine the velocity at different distances from the starting point. However, you need to separate the calculations of velocity with respect to distance into those above the starting point and those below, because those above the starting point have two solutions—upward and downward.
Questions you may have include:
- What the velocity for given time?
- What is the velocity for given distance above the starting point?
- What is the velocity for given distance below the starting point?
This lesson will answer those questions. There is a mini-quiz near the end of the lesson.
Useful tools: Metric-English Conversion | Scientific Calculator.
Velocity for a given time
The equation for the velocity of an object thrown upward after a given amount of time is:
v = gt + vi
where
- v is the velocity 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)
- vi is the upward initial velocity in m/s or ft/s (a negative number)
Positive and negative velocities
Since the velocity of an object falling toward the ground is considered a positive number, the velocity going upward is considered a negative number. Thus, for an object projected upward, the initial velocity vi is a negative number.
As the object goes upward, its velocity v is also negative. Once the object starts moving downward, v is positive.
Note: Some textbooks consider up as positive and down as negative. We feel it is more logical to consider an object that is accelerating downward as a positive velocity. However, you need to be aware of what convention is being used when working from a book.
Obviously, the time t is always a positive number.
Example
If the initial velocity of an object is 19.6 m/s in the upward direction, what is the velocity are various times?
Solution
Since vi is in m/s, g = 9.8 m/s2. Substitute for vi and g in the equation:
v = gt + vi
v = (9.8 m/s2)(t s) + (−19.6 m/s)
or simply
v = (9.8t − 19.6) m/s
Thus, for various values of t, you can calculate v:
t = 0 s v = −19.6 m/s Moving upward from starting point t = 1 s v = −9.8 m/s Object moving upward t = 2 s v = 0 m/s At peak or maximum height t = 3 s v = 9.8 m/s Object moving downward t = 4 s v = 19.6 m/s Passing starting point t = 5 s v = 29.4 m/s Continuing downward
Velocities for different times of object projected upward
Velocity for distances above starting point
The equation for the velocity of an object thrown upward after it has traveled a given distance above the starting point is:
v = ±√(2gx + vi2)
where
- ± means plus or minus
- √(2xg + vi2) is the square root of the quantity (2xg + vi2)
- x is the distance above the starting point in m or ft (a negative number)
The equation shows that the velocity for distances above the starting point can be either negative (−) or positive (+), depending on whether the object is going up (negative) or down (positive).
Since x is a negative number, it cannot have a value such that (2xg + vi2) is a negative number. This is because is not possible to take the square root of a negative number.
Example above starting point
If vi = −64 ft/s, find the values of v for various distances x above the starting point.
Solution
Since g = 32 ft/s2, substitute for g and vi in the equation:
v = ±√(2gx + vi2)
v = ±√[2*(32 ft/s2)*(x ft) + (−64 ft/s)2]
v = ±√(64x ft2/s2 + 4096 ft2/s2)
Since √( ft2/s2) = ft/s, you get:
v = ±√(64x + 4096) ft/s
Substituting values for x above the starting point, allows you to calculate values for v:
x = 0 ft v = −64 ft/s Moving upward from starting point x = −32 ft v = −45.3 ft/s Object moving upward x = −64 ft v = 0 ft/s At peak or maximum height x = −32 ft v = +45.3 ft/s Object moving downward x = −64 ft v = +64 ft/s Passing downward at starting point
Velocities for distances of object projected upward above starting point
Velocity for distances below starting point
Below the starting point, both x and v are positive numbers, while vi is still
a negative number.
v = +√(2gx + vi2)
Example below starting point
If vi = −20 m/s and g = 9.8 m/s2, then the equation is:
v = √[2*(9.8 m/s2)(x m) + (−20 m/s)2]
v = √(19.6x m2/s2 + 400 m2/s2)
or simply
v = √(19.6x + 400) m/s
Thus, when x equals certain values, you can calculate v:
x = 1 m v = 20.5 m/s x = 10 m v = 24.4 m/s x = 20 m v = 28.1 m/s
Velocities for distances of object projected upward below starting point
Summary
When an object is thrown upward, the initial velocity is a negative number. The velocity is negative while the object moves up and positive while it moves downward. The distance is negative above the starting point positive below the starting point.
The equations for velocity are:
v = gt + vi
v = ±√(2gx + vi2) above the starting point
v = √(2gx + vi2) below the starting point
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Resources
The following resources provide information on this subject:
Websites
Acceleration due to Gravity Calculations - from Western Washington University
Gravitation and Gravity 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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Gravity Velocity Equations for Objects Projected Upward