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Gravity Distance Equations for Objects Projected Upward
by Ron Kurtus (15 August 2009)
If you know the initial velocity that an object is thrown upward, you can calculate the distance traveled from the starting point when moving at a given velocity. You can also determine the velocity at different distances from the starting point.
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 is the distance for a given velocity?
- How do you find the maximum height of a projected object?
- What is the distance for a given time?
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 to reach a given velocity
The equation for the distance an object thrown upward travels to reach a given velocity is according to the equation:
x = (v2 − vi2)/2g
where
- x is the distance traveled from the starting point in meters (m) or feet (ft)
- v is the velocity in m or ft
- vi is the initial velocity in m or ft (a negative number)
- g is the acceleration due to gravity (9.8 m/s2 or 32 ft/s2)
Positive and negative values
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.
The distance from the starting point to the object when it is above the starting point is considered a negative value. Once the object is below the starting point, the distance 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 1
Suppose you throw a ball upward at 10 feet per second. How far does the ball travel from the starting point to reach a velocity of 4 ft/s in an upward direction?
Solution
Since vi and v are in ft/s, g = 32 ft/s2. Also, vi = −10 ft/s and v = −4 ft/s, since we consider upward as a negative direction. Substitute values for v, vi and g into the equation:
x = (v2 − vi2)/2g
x = [(−4 ft/s)2 − (−10 ft/s)2]/2*(32 ft/s2)
Note: The square of a negative number is positive: (−4)*(−4) = +16
x = (16 ft2/s2 − 100 ft2/s2)/(64 ft/s2)
x = (−84 ft2/s2)/(64 ft/s2)
x = −1.3125 ft
That is a little more than 1.3 feet above from the starting point.
Example 2
How far does this ball travel from the starting point to reach a velocity of 20 ft/s in the downward direction?
Solution
Again, since vi and v are in ft/s, g = 32 ft/s2. vi = −10 ft/s and v = 20 ft/s. Substitute values for v, vi and g into the equation:
x = (v2 − vi2)/2g
x = [(20 ft/s)2 − (−10 ft/s)2]/2*(32 ft/s2)
x = (400 ft2/s2− 100 ft2/s2)/(64 ft/s2)
x = (300 ft2/s2)/(64 ft/s2)
x = 4.69 feet below the starting point
Distance to maximum height
The distance an object thrown upward travels to reach its peak or maximum height before descending requires that the velocity v = 0. The equation is:
xmax = −vi2/2g
where xmax is the maximum height or distance.
Since the peak of the travel is above the starting point, xmax is a negative number. To avoid confusion, it is good to state xmax as a positive number above the starting point.
Example 1
If the initial velocity is vi = −10 ft/s, what is the maximum height?
Solution
Since vi is in ft/s, g = 32 ft/s2. Substitute for vi and g into the equation:
xmax = −vi2/2g
xmax = −(−10 ft/s)2]/2*(32 ft/s2)
xmax = −(100 ft2/s2)/(64 ft/s2)
xmax = −1.56 ft or 1.56 ft above the starting point
Example 2
If the initial velocity is vi = −20 m/s, what is the maximum height?
Solution
Since vi is in m/s, g = 9.8 m/s2. Substitute for vi and g into the equation:
xmax = −vi2/2g
xmax = −(−20 m/s)2]/2*(9.8 m/s2)
xmax = −(400 m2/s2)/(19.6 m/s2)
xmax = −20.4 m or 20.4 above the starting point
Distance in a given time
The distance an object thrown upward travels within a given time is according to the equation:
x = gt2/2 + vit
where t is the time in seconds (s).
Distance for various times
If the initial velocity is vi = −20 m/s, then g is 9.8 m/s2 and the equation becomes:
x = 9.8t2/2 − 20t
x = (9.8 m/s2)*(t2 s2)/2 + (−20 m/s)*(t s)
x = (4.9t2 − 20t) m
Substituting values for t, you get values for x:
t = 0 s x = 0 m At the starting point t = 1 s x = −15.1 m Above starting point t = 2 s x = −20.4 m Close to peak t = 3 s x = −15.9 m Falling but above starting point t = 4 s x = −1.6 m Nearing starting point t = 5 s x = 22.5 m 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 and positive below the starting point.
The equations for distance are:
x = (v2 − vi2)/2g
xmax = −vi2/2g
x = gt2/2 + vit
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Acceleration due to Gravity Calculations - from Western Washington University
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Mini-quiz to check your understanding
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Gravity Distance Equations for Objects Projected Upward