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Gravity Velocity Equations for Objects Projected Upward
by Ron Kurtus (revised 11 August 2010)
When you project an object upward, it has an initial velocity less than zero
(vi < 0). The object slows down as it moves upward until it reaches a maximum height, at which time
the velocity is zero. Then the velocity increases as the object falls toward the ground.
Derived equations allow you to calculate the velocity of an object projected upward with respect to time, as well as distance both above and below the starting point.
Note: The convention we use is that upward velocities are negative and downward velocities are positive. Also, distances above the starting point are negative and those below the starting point are positive.
Some textbooks consider up as positive and down as negative. You need to be aware of what convention is being used when working from a book.
(See Convention for Gravity Direction for more information.)
Questions you may have include:
- What the velocity with respect to time?
- What is the velocity for distances above the starting point?
- What is the velocity for distances 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 with respect to time
The equation for the velocity of an object thrown upward with respect to time is:
v = gt + vi
where
- v is the vertical 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 vertical velocity in m/s or ft/s
(See Derivation of Velocity-Time Gravity Equations for more information.)
Since the initial velocity is negative or less than zero (vi < 0), the various values for time t determine whether the object is moving upward (v < 0), at the maximum height (v = 0) or moving downward (v > 0).
Example
If the initial velocity of an object is 19.6 m/s in the upward direction, what is the velocity at various times?
Solution
vi = −19.6 m/s and g = 9.8 m/s2. Substitute for vi and g in the equation in order to get a formula in terms of t:
v = gt + vi
v = (9.8 m/s2)(t s) + (−19.6 m/s)
Simplify:
v = (9.8t − 19.6) m/s
Substitute values of t in the formula:
t = 0 s v = −19.6 m/s Moving upward from starting point t = 1 s v = −9.8 m/s Object moving upward tm = 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 general equation for the velocity of an object projected upward with respect to the distance above the starting point is:
v = ±√(2gy + vi2)
where
- ± means plus or minus
- √(2yg + vi2) is the square root of the quantity (2yg + vi2)
- y is the vertical distance above the starting point in m or ft (y < 0)
(See Derivation of Distance-Velocity Gravity Equations for more information.)
Since you know that v < 0 on the way up and v > 0 on the way down, you can break the equation into its up and down components:
v = −√(2gy + vi2) (on the way up)
v = √(2gy + vi2) (on the way down)
Example
If vi = −64 ft/s, find the values of v for various distances y above the starting point.
Solution
Since g = 32 ft/s2, substitute for g and vi in the equation to get a formula in terms of y:
v = ±√(2gy + vi2)
v = ±√[2*(32 ft/s2)*(y ft) + (−64 ft/s)2]
v = ±√(64y ft2/s2 + 4096 ft2/s2)
Since √( ft2/s2) = ft/s, you get:
v = ±√(64y + 4096) ft/s
Substituting values for y in the formula, knowing that the negative value is on the way up and the positive value is on the way down:
y = 0 ft v = −64 ft/s Moving upward from starting point y = −32 ft v = −45.3 ft/s Object moving upward ym = −64 ft v = 0 ft/s At peak or maximum height y = −32 ft v = +45.3 ft/s Object moving downward y = −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 y and v are positive numbers, while vi is still
a negative number.
v = √(2gy + vi2)
Example
If vi = −20 m/s and g = 9.8 m/s2, what are the velocities for various distances?
Solution
Substitute for vi and g in the equation:
v = +√(2gy + vi2)
v = √[2*(9.8 m/s2)(y m) + (−20 m/s)2]
v = √(19.6y m2/s2 + 400 m2/s2)
Simplify:
v = √(19.6y + 400) m/s
Substitute values for y in the formula:
y = 1 m v = 20.5 m/s y = 10 m v = 24.4 m/s y = 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 = ±√(2gy + vi2) above the starting point
v = √(2gy + 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
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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Gravity Velocity Equations for Objects Projected Upward