Gravity Time Equations for Objects Projected Upward

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Gravity Time Equations for Objects Projected Upward

by Ron Kurtus (revised 8 August 2010)

When you project an object upward, it has an initial velocity less than zero
(v_i < 0). The object slows down until it reaches a maximum height, after which it increases its velocity as it falls toward the ground.

Derived equations allow you to calculate the time it takes the object to reach a given velocity, as well as the time to reach a given distance from 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. Obviously, time is always 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 time with respect to velocity?
What is the time for distances above the starting point?
What is the time 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.

Time with respect to velocity

The equation for the time it takes for the object that is projected upward to reach a given velocity is:

t = (v − v_i)/g

where

t is the time in seconds (s)
v is the vertical velocity in meters/second (m/s) or feet/second (ft/s)
v_i is the vertical initial velocity in m/s or ft/s
g is the acceleration due to gravity (9.8 m/s² or 32 ft/s²)

(See Derivation of Velocity-Time Gravity Equations for more information.)

Time to reach maximum height

At the maximum height, v = 0 and the equation becomes:

t_m = −v_i/g

where t_m is the time to reach the maximum height.

Note: Since v_i < 0, −v_i is a positive number.

Example of times to reach various velocities

If v_i = −128 ft/s, find t for various velocities.

Solution

Since v_i is in ft/s, g = 32 ft/s². Substitute values for v_i and g into the equation:

t = (v − v_i)/g

t = [(v ft/s) − (−128 ft/s)]/(32 ft/s²)

Cancelling out units results in the formula:

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_m = 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

Times for various velocities of object projected upward

Time for distances moving upward

The equation for the time it takes an object to move upward toward the maximum height is:

t = [−v_i − √(v_i²+ 2gy)]/g

where

√(v_i²+ 2gy) is the square root of the quantity (v_i² + 2gy)
y is the vertical distance from the starting point in meters or feet (y < 0)

(See Derivation of Distance-Time Gravity Equations for more information.)

Maximum height

The time it takes to reach the maximum height is:

t_m = √(−y_m/g)

where y_m is the maximum height or distance above the starting point. According to our convention for directions, y_m < 0.

It is convenient to know the relationships between the time it takes to reach the maximum height and initial velocity, as well as the maximum height and initial velocity:

t_m = −v_i/g

y_m = −v_i²/2g

Example for various distances moving upward

If v_i = −98 m/s, find the times for various distances as the object moves upward to the maximum height.

Solution

Since v_i is in m/s, g = 9.8 m/s². Also, the values of y will be negative numbers.

You can easily determine the time for the maximum height:

t_m = −v_i/g

t_m = 98/9.8 s

t_m = 10 s

Also:

y_m = −v_i²/2g

y_m = −98²/2*9.8 m

y_m = −980/2 m

y_m = −490 m

Substitute values for v_i and g into the equation:

t = [−v_i − √(v_i²+ 2gy)]/g

You can independently verify that the units are correct and that t is in seconds.

t = [−(−98) − √(−98² + 2*9.8*y)]/(9.8) seconds

One way to simplify the equation is by breaking up the fraction.

t = 98/9.8 − [√(9604 + 19.6y)]/(9.8) s

t = 10 − [√(9604 + 19.6y)]/9.8 s

The equation still is not very simple, but it is workable. A simple case is when
y = 0:

t = 10 − [√(9604)]/9.8 s

t = 10 − 98/9.8 s

t = 0 s

The following results show the times for various distances above and at the starting point:

y = 0 m t = 0 s At the starting point

y = −49 m t = 0.5 s Moving up and above starting point

y = −98 m t = 1.1 s Moving upward

y = −196 m t = 2.25 s Moving upward

y_m = −490 m t_m = 10 s At maximum height

Times for various distances in upward direction

Time for distances moving downward

After the object reaches the maximum height and starts coming down, the equation changes to:

t = [−v_i + √(v_i²+ 2gy)]/g

When the object is above the starting point, y is still negative (y < 0). When it reaches the starting point, y = 0. Then, below the starting point, y > 0.

Example for various distances moving downward

As in the previous example, if v_i = −98 m/s. Find the times for various distances as the object moves downward.

Solution

You already know the maximum height and time:

y_m = −490 m

t_m = 10 s

Use the equation:

t = [−v_i + √(v_i²+ 2gy)]/g

Substitute and simplify:

t = 10 + [√(9604 + 19.6y)]/9.8 s

The following results show the times for various distances as the object falls from the maximum height:

y_m = −490 m t_m = 10 s At maximum height

y = −196 m t = 17.75 Falling

y = −98 m t = 18.9 s Falling but above starting point

y = −49 m t = 19.5 s Nearing starting point on way down

y = 0 m t = 20 s At starting point on the way down

y = 196 m t = 21.8 s Below the starting point

Times for various distances in downward direction

Summary

An object projected upward against the force of gravity slows down until it reaches a maximum height, after which it increases its velocity as it falls toward the ground.

The equations below allow you to calculate the time it takes an object projected upward to reach a given velocity, as well as a given distance from the starting point:

t = (v − v_i)/g

t = [−v_i − √(v_i²+ 2gy)]/g (moving upward)

t_m = √(−y_m/g) (time to reach maximum height)

t = [−v_i + √(v_i²+ 2gy)]/g (moving downward)

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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_m = 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

y = 0 m	t = 0 s	At the starting point
y = −49 m	t = 0.5 s	Moving up and above starting point
y = −98 m	t = 1.1 s	Moving upward
y = −196 m	t = 2.25 s	Moving upward
y_m = −490 m	t_m = 10 s	At maximum height

Gravity and Gravitation

Gravity topics

Derivations of equations

Falling objects

Thrown downward

Thrown upward

Gravity applications

Gravitation topics

Theories

Equations

Gravity applications

Escape velocity

Time with respect to velocity

Time to reach maximum height

Example of times to reach various velocities

Solution

Time for distances moving upward

Maximum height

Example for various distances moving upward

Solution

Time for distances moving downward

Example for various distances moving downward

Solution

Summary

See the Side Menu for more Gravity and Gravitation topics

Resources

Websites

Books

Mini-quiz to check your understanding

What do you think?

Share link

Students and researchers

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Gravity Time Equations for Objects Projected Upward

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