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Gravity Distance Equations for Objects Projected Upward
by Ron Kurtus (revised 11 August 2010)
When you throw or project an object upward, it has an initial velocity less than zero (vi < 0). The object moves upward until it reaches a maximum height, after which it falls toward the ground.
Derived equations allow you to calculate the distance the object travels from the starting point with respect to velocity, as well as with respect to time for both above and below the starting point. You can also calculate the the total distance the object travels, both up and down.
Note: The convention we use is that distances above the starting point are negative and those below the starting point are positive. Also, an upward velocity is negative and downward velocity is 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 is the distance with respect to velocity?
- What is the distance with respect to time?
- What is the total distance traveled?
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 with respect to velocity
The equation for the distance from the starting point of an object projected upward with respect to velocity is:
y = (v2 − vi2)/2g
where
- y is the vertical distance traveled from the starting point in meters (m) or feet (ft)
- v is the vertical velocity in m/s or ft/s
- vi is the initial vertical velocity in m/s or ft/s; vi < 0
- 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.)
Moving upward, above starting point
When the object is moving upward, the velocity is a negative number (v < 0). The vertical distance from the starting point is also a negative number (y < 0).
At maximum height
At the peak or maximum height, the velocity is v = 0 and the distance is:
ym = −vi2/2g
where ym is the maximum height or distance. Since the peak of the travel is above the starting point, ym is a negative number.
Moving downward
When the object is moving downward from the peak height, the velocity is then a positive value (v > 0).
The distance from the starting point is still negative (y < 0) when the object is above the starting point. However, once the object falls below the starting point, the distance becomes a positive number (y > 0).
Example
Suppose you throw a ball upward at 100 feet per second. What are the distances from the starting point for the various velocities?
Solution
Since vi = −64 ft/s, g = 32 ft/s2. Substitute values for vi and g into the equation:
y = (v2 − vi2)/2g
y = [v2 ft2/s2 − (−64 ft/s)2]/2*(32 ft/s2)
Combine and cancel out the units to get the formula:
y = (v2 − 4096)/(64) ft
Substitute in values for v:
v = −64 ft/s y = 0 ft At the starting point v = −32 ft/s y = −48 ft Above starting point v = 0 ft/s ym = −64 ft Maximum height v = +32 ft/s y = −48 ft Falling but above starting point v = +64 ft/s y = 0 ft At the starting point v = +80 ft/s y = +36 ft Below starting point
Distances for various velocities of object projected upward
Distance with respect to time
The equation for the distance an object thrown upward travels within a given time period is:
y = gt2/2 + vit
where t is the time in seconds (s).
(See Derivation of Distance-Time Gravity Equations for details of the derivations.)
Time to maximum height
The time it takes the object to reach the maximum height ym = −vi2/2g is:
tm = −vi/g
Example
If the initial velocity is vi = −20 m/s, what are the distances for various times?
Solution
Substitute values for viand g in the equation:
y = gt2/2 + vit
y = (9.8 m/s2)*(t2 s2)/2 + (−20 m/s)*(t s)
Combine and cancel out units to get the formula:
y = (4.9t2 − 20t) m
Substitute in values for t. But also note that tm = −vi/g and ym = −vi2/2g:
tm = 20/9.8 s = 2.04 s
ym = −400/19.6 m = 20.4 m
t = 0 s y = 0 m At the starting point t = 1 s y = −15.1 m Above starting point tm = 2 s ym = −20.4 m At maximum height t = 3 s y = −15.9 m Falling but above starting point t = 4 s y = −1.6 m Nearing starting point t = 5 s y = 22.5 m Below starting point
Distances for various times of object projected upward
Total distance traveled
The distance y in the stated equations is the distance from the starting point, with y < 0 above the starting point and y > 0 below the point. In order to find the total distance traveled, you need to state where the object is and then add or subtract the various elements. It is necessary to know the maximum height, thus making calculation of the total distance with respect to time difficult.
Note: Total distance is always a positive number. It is not related to direction.
Total distance going up
The total distance traveled for an object moving upward is simply absolute value of the calculated distance from the starting point:
du = |y|
where
- du is the total distance going up
- |y| is the absolute value of the distance from the starting point
Note: The absolute value of y is designated as |y|, which means the positive value of y.
Example
In the previous example, where vi = −20 m/s, the distance traveled at
t = 1 second is y = −15.1 m. Thus, the total distance is:
du = |−15.1|
du = 15.1 m
Total distance coming down, above starting point
The total distance the object travels on the way down, when above the starting point, is the absolute value of the sum of the two times maximum height and the distance from the starting point:
dd = |2ym + y|
where
- dd is the total distance coming down above the starting point
- ym = −vi2/2g.
Example
When t = 3 s, y = −15.9 m. Also, ym = −20.4 m
dd = |2ym + y|
dd = |−20.4 + (−15.9)|
dd = |−36.3|
dd = 36.3 m
Total distance below the starting point
The total distance below the starting point is two times the absolute value to the maximum height, plus the distance below the starting point.
db = |2ym| + y
where db is the total distance below the starting point.
Example
When t = 5 s, y = 22.5 m. Also ym = −20.4 m:
db = |2ym| + y
db = |−40.8| + 22.5
db = 40.8 + 22.5
db = 63.3 m
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 and positive below the starting point.
The equations for distance are:
y = (v2 − vi2)/2g
ym = −vi2/2g (maximum height)
y = gt2/2 + vit
The total distance traveled is:
du = |y| (going upward)
dd = |2ym + y| (going downward above starting point)
d = |2ym| + y (below 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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