# Gravity Displacement Equations for Objects Projected Upward

by Ron Kurtus (revised 7 February 2011)

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

Since it is moving in the opposite direction of the force of gravity, its initial velocity is a negative number. Also, according to our convention for direction, the displacement above the starting point is negative, while the displacement below the starting point is positive.

Note: Some textbooks consider upward motion as positive and downward as negative. You need to be aware of what convention is being used when working from a book.

Derived equations allow you to calculate the displacement with respect to velocity and with respect to elapsed time. You can also calculate the total distance the object travels from the starting point.

Questions you may have include:

• What is the displacement with respect to velocity?
• What is the displacement with respect to time?
• What is the total distance traveled?

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

## Displacement with respect to velocity

The general gravity equation for the displacement of an object with respect to velocity is:

y = (v2 − vi2)/2g

where

• y is the vertical displacement 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
• g is the acceleration due to gravity (9.8 m/s2 or 32 ft/s2)

(See Derivation of Displacement-Velocity Gravity Equations for details of the derivation.)

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

While the object is moving upward, the square of its velocity is less than the square of the initial velocity (v2 < vi2). The result is the displacement from the starting point is negative (y < 0).

### Maximum displacement with respect to velocity

At the peak or maximum displacement, the velocity is v = 0 and the displacement is:

ym = (0 − vi2)/2g

ym = −vi2/2g

where ym is the maximum displacement.

### On the way down

When the object is moving downward from the peak displacement but above the starting point, v2 is still less than vi2 and the displacement is still negative (y < 0).

Once the object falls below the starting point, (v2 > vi2) and the displacement becomes a positive value (y > 0).

### Example

Suppose you throw a ball upward at 100 feet per second. What are the displacements 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 displacement 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

## Displacement with respect to time

The general gravity equation for the displacement of an object with respect to time is:

y = gt2/2 + vit

where t is the time in seconds (s).

(See Derivation of Displacement-Time Gravity Equations for details of the derivation.)

Since the initial velocity is negative, y will be negative for values of t when:

gt2/2 < |vi|t

where |vi| is the absolute or positive value of vi. These are values of t where the object is above the starting point.

When gt2/2 > |vi|t, the displacement y is positive and the object is below the starting point.

### Maximum displacement with respect to time

The equation for the maximum displacement with respect to time can be determined by starting with the equation:

tm = −vi/g

where tm is the time to reach the maximum displacement

Solve for vi:

vi = −gtm

Substitute for vi in y = gt2/2 + vit:

ym = gtm2/2 − gtm2

ym = −gtm2/2

### Example

If the initial velocity is vi = −20 m/s, what are the displacements 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 displacement 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 Displacements for various times of object projected upward

## Total distance traveled

The displacement of an object is its movement in a specific direction from a starting point to some end point. It is a vector quantity. Distance is a scalar quantity that is independent of direction and always has a positive value.

In the stated equations, the displacement y is negative above the starting point, when the object is going upward and downward. The value for y is positive below the starting point.

In order to find the total distance traveled, you need to state where the object is and then add the various displacements. In some cases, it is necessary to also know the maximum displacement.

### Distance going up

The distance traveled of an object moving upward is simply the absolute value of the object's displacement:

du = |y|

where

• du is the distance going up
• |y| is the absolute or positive value of the displacement going up

At the maximum displacement:

dm = |ym|

### Total distance going up plus coming down

The total distance the object travels going up and then coming down, is the absolute value of two times the distance to the maximum displacement plus the displacement to the end point:

d = |2ym| + y

where

• d is the total distance going up plus coming down
• ym = −vi2/2g

If the object is above the starting point, y is negative and subtracted from |2ym|. Below the starting point, everything is positive.

### Example

Find the total distance traveled when the displacement is y = −5 m, ym = −10 m and y = +5 m. Total distances of object projected upward

#### Going up

The total distance going up is:

du = |−5| m

du = 5 m

#### At maximum displacement

The total distance at the maximum displacement is:

dm = |−10| m

dm = 10 m

#### Coming down, above starting point

The total distance coming down and above the starting point is:

d = |−20| + (−5) m

d = 20 5 m

d = 15 m

#### Coming down, below starting point

The total distance coming down and below the starting point is:

d = |−20| + (+5) m

d = 20 + 5 m

d = 25 m

## Summary

When you project an object upward and release it at some initial velocity, it travels upward until it reaches a maximum displacement, after which it falls toward the ground. The displacement above the starting point is negative, while the displacement below the starting point is positive.

Derived equations allow you to calculate the displacement for a given velocity, as well as the elapsed time for displacements both above and below the starting point. You can also calculate the total distance the object travels from the starting point. The equations for displacement are:

### With respect to velocity

y = (v2 − vi2)/2g

ym = −vi2/2g (maximum displacement)

### With respect to time

y = gt2/2 + vit

ym = −gtm2/2 (maximum displacement)

### Total distance traveled

du = |y| (going upward to maximum displacement)

dd = |2ym| + y (sum of going up and coming down)

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