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Gravity Time Equations for Objects Projected Upward

by Ron Kurtus (revised 7 February 2011)

When you project an object upward and release it at some initial velocity, it travels until it reaches a maximum height or displacement, after which it falls toward the ground.

The initial velocity has a negative value, the velocity is negative on the way up and positive on the way down, and the displacement is negative above the starting point and positive below the starting point. Velocity and displacement are vectors, while time is a scalar quantity, which is always positive.

Note: 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 Direction in Gravity Equations for more information.)

You can use derived equations to find the time it takes to reach a given velocity, as well as the time it takes to reach a given displacement, both above and below the starting point.

Questions you may have include:

This lesson will answer those questions. Useful tool: Units Conversion



Time with respect to velocity

The general equation for the time it takes for an object to reach a given velocity is:

t = (v − vi)/g

where

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

When you project the object upward, it is moving in the opposite direction of gravity, and the initial velocity when you release it is negative or less than zero
(vi < 0).

On the way up

On the way up, v is negative (v < 0), and its absolute value is less than that of vi:

|v| < |vi|

where |v| and |vi| are the absolute or positive values. Thus, (v − vi) has a positive value.

Time to reach maximum displacement

At the maximum displacement, v = 0 and the time equation becomes:

tm = −vi/g

where tm is the time to reach the maximum displacement.

Note: Since the initial velocity is negative, −vi is a positive number.

On the way down

On the way down, v > 0 and v < |vi|. Thus, (v − vi) has a positive value.

Example of times to reach various velocities

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

Solution

Since vi is in ft/s, you use 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)

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 tm = 4 s At maximum displacement
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

Times for various velocities of object projected upward

Time for displacements moving upward

The general gravity equation for elapsed time with respect to displacement is:

t = [ −vi ± √(vi2 + 2gy) ]/g

where

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

For an object projected upward, −vi is a positive number. In order for the object to start at t = 0 when y = 0, the equation for the time it takes an object to move upward toward the maximum displacement is the negative (−) version:

t = [−vi − √(vi2 + 2gy)]/g

Time to reach maximum displacement

The equation for the time to reach maximum displacement with respect to the initial velocity has already been stated:

tm = −vi/g

The equation for the maximum displacement with respect to the initial velocity is:

ym = −vi2/2g

(See Displacement Equations for Objects Projected Upward for more information.)

These two equations make it easy to determine the time and maximum displacement. However, you may want to see the relationship between the factors.

Solve ym = −vi2/2g for vi2:

vi2 = −2gym

Square tm = −vi/g:

tm2 = vi2/g2

Substitute in vi2 = −2gym:

tm2 = −2gym/g2

tm2 = −2ym/g

Take positive square root:

tm = √(−2ym/g)

Example for various displacements moving upward

If vi = −98 m/s, find the times for various displacements as the object moves upward to the maximum displacement.

Solution

Since vi is in m/s, g = 9.8 m/s2. Also, since the displacements are above the starting point, the values of y will be negative numbers.

You can easily determine the time for the maximum displacement:

tm = −vi/g

tm = 98/9.8 s

tm = 10 s

Also:

ym = −vi2/2g

ym = −982/2*9.8 m

ym = −980/2 m

ym = −490 m

You can verify those two values are correct by substituting in:

tm = √(−2ym/g)

10 = √(2*[490]/9.8)

10 = √(100)

10 = 10

To find the various values of t with respect to y, substitute for vi and g into the equation:

t = [−vi − √(vi2 + 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 displacements 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
ym = −490 m tm = 10 s At maximum displacement

Times for various displacements in upward direction

Times for various displacements in upward direction

Time for displacements moving downward

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

t = [−vi + √(vi2 + 2gy)]/g

While the object is above the starting point, y is negative. When it reaches the starting point, y = 0. Then, when it is below the starting point, y becomes a positive number.

Example for various displacements moving downward

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

Solution

You already know the maximum displacement and time:

ym = −490 m

tm = 10 s

Use the equation:

t = [−vi + √(vi2 + 2gy)]/g

Substitute and simplify:

t = 10 + [√(9604 + 19.6y)]/9.8 s

The following results show the times for various displacements as the object falls from the maximum displacement:

ym = −490 m tm = 10 s At maximum displacement
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 displacements in downward direction

Times for various displacements in downward direction

Summary

An object projected upward against the force of gravity slows down until it reaches a maximum displacement, after which its velocity increases as it falls toward the ground. Derived equations allow you to calculate the time it takes an object projected upward to reach a given velocity or a given displacement from the starting point.

Time with respect to velocity:

t = (v − vi)/g

tm = −vi/g (time to reach maximum displacement)

Time with respect to displacement:

t = [−vi − √(vi2 + 2gy)]/g (moving upward)

tm = √(−2ym/g) (time to reach maximum displacement)

t = [−vi + √(vi2 + 2gy)]/g (moving downward)


Be methodical in your calculations


Resources and references

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