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Gravity Time Equations for Objects Projected Upward
by Ron Kurtus (revised 20 October 2009)
If you know the initial velocity at which an object is thrown upward, you can calculate the time it takes for the object to reach a given velocity or for it to travel a certain distance from the starting point. You need to separate the calculations of time 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 time to reach a given velocity?
- What is the time to reach a distance above the starting point?
- What is the time to reach a 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.
Time to reach a given velocity
The equation for the time it takes for the object that is projected upward to reach a given velocity is:
t = (v − vi)/g
where
- t is the time in seconds (s)
- v is the velocity in meters/second (m/s) or feet/second (ft/s)
- vi is the initial velocity in m/s or ft/s (negative number)
- g is the acceleration due to gravity (9.8 m/s2 or 32 ft/s2)
Positive and negative numbers
Since going toward the ground is considered a positive number, v is a negative number when the object is going upward. Keep the direction in mind when defining values for v in the equation.
Likewise, the initial velocity is upward and is a negative number. Also, since vi is a negative number, −vi is a positive number.
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 vi = −128 ft/s, find t for various velocities.
Solution
Since vi is in ft/s, g = 32 ft/s2. Substitute values for vi and g into the equation:
t = (v − vi)/g
t = [(v ft/s) − (−128 ft/s)]/(32 ft/s2)
Since (ft/s)/(ft/s2) = (ft/s)*(s2/ft) = seconds, you get:
t = (v + 128)/32 s
Substitute different values for v to get the various elapsed times:
v = −128 ft/s t = 0 s Starting off at initial velocity v = −64 ft/s t = 2 s Moving upward v = 0 ft/s t = 4 s At maximum height or peak v = 32 ft/s t = 5 s Moving downward v = 128 ft/s t = 8 s Downward at starting point v = 160 ft/s t = 9 s Below starting point
Time to reach a distance above starting point
The equation for the time it takes an object projected upward to reach a given distance above the starting point is:
−vi ± √(vi2 + 2gx)
t = ________________
g
The equation can also be written in a more compact form:
t = [−vi ± √(vi2 + 2gx)]/g
where
- ± means plus or minus
- √(vi2 + 2gx) is the square root of the quantity (vi2 + 2gx)
- x is the distance from the starting point in meters or feet (a negative number)
This equation means that for a given distance above the starting point, there are two possible values for t. One value is the time is takes to reach the distance, while going up, and the other value for t is the time it takes to get to the same position on the way down.
Positive and negative values
You need to be aware of the positive and negative values when solving the equation.
Since vi is a negative number, −vi is a positive number.
The distance, x, is a negative number above the starting point. The distance below the starting point is a positive number.
Since the time is a positive number, −vi (a positive number) must be greater than −√(vi2+ 2gx). This is true only when x is a negative number or the object is above the starting point.
Example
If vi = −98 m/s, find the times for various distances above the starting point. (This means that the values of x will be negative numbers.)
Solution
Since vi is in m/s, g = 9.8 m/s2. Substitute values for vi and g into the equation:
−vi ± √(vi2 + 2gx)
t = ________________
g
−(−98 m/s) ± √[(−98 m/s)2 + 2*(9.8 m/s2)*(x m)]
t = ________________________________________
(9.8 m/s2)
Or you can substitute into:
t = [−vi ± √(vi2 + 2gx)]/g
t = {−(−98 m/s) ± √[(−98 m/s)2 + 2*(9.8 m/s2)*(x m)]}/(9.8 m/s2)
Either equation is horrendous, when the units are included. However, you need to make sure the units cancel out, such that t is in seconds.
t = [(98) ± √(982 + 2*9.8x)]/(9.8) seconds
One way to simplify the equation is by breaking up the fraction.
t = 98/9.8 ± [√(9604 + 19.6x)]/(9.8) s
t = 10 ± [√(9604 + 19.6x)]/9.8 s
It still is not very simple, but it is workable. The following results show the times for various distances above and at the starting point.
x = 0 m t = 0 s At the starting point x = −49 m t = 0.5 s Moving up and above starting point x = −98 m t = 1.1 s Moving upward x = −196 m t = 2.25 s Moving upward x = −490 m t = 10 s At maximum height x = −196 m t = 17.75 Falling x = −98 m t = 18.9 s Falling but above starting point x = −49 m t = 19.5 s Nearing starting point on way down x = 0 m t = 20 s At starting point on the way down
Time to reach distance below starting point
When an object projected upward passes below the starting point, the value of the distance is a positive number. The time it takes an object projected upward to reach a given distance below the starting point is simply:
t = [−vi + √(vi2 + 2gx)]/g
Continuing from the chart above with vi = −98 m/s and x = +196 m:
t = 10 + [√(9604 + 19.6*196)]/9.8 s
t = 10 + [√(9604 + 19.6*196)]/9.8 s
t = 10 + [√(9604 + 3841.6)]/9.8 s
t = 10 + (√13445.6)/9.8 s
t = 10 + 116/9.8 s
t = 21.8 s
Certainly, the calculations are tedious.
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 time are:
t = (v − vi)/g
t = [−vi ± √(vi2 + 2gx)]/g (above the starting point)
t = [−vi + √(vi2 + 2gx)]/g (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 Time Equations for Objects Projected Upward