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Gravity Velocity Equations for Objects Projected Upward

by Ron Kurtus (revised 6 February 2011)

When you project an object upward and release it at its initial velocity, it is moving in the opposite direction of the force of gravity. Thus the initial velocity is negative. The velocity of the object is also negative on the way up but positive on the way down.

Note: The convention we use is that upward velocities are negative and downward velocities are positive. Also, displacements above the starting point are negative and those below the starting point are 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 Direction in Gravity Equations for more information.)

The object slows down as it moves upward until it reaches a maximum height, at which time the velocity is zero. Then the velocity increases as the object falls toward the ground.

Derived equations allow you to calculate the velocity of an object projected upward with respect to time, as well as the velocity at displacements both above and below the starting point.

Questions you may have include:

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



Velocity with respect to time

The general gravity equation for the velocity with respect to time is:

v = gt + vi

where

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

When you project the object upward, the initial velocity when you release it is negative or less than zero (vi < 0). The resulting velocities will be negative (v < 0) when the object is moving upward, zero (v = 0) at the maximum displacement or positive (v > 0) when the object is moving downward, depending on the value for elapsed time.

Time to maximum displacement

When determining the velocity at various times, it is convenient to know the time required to reach the maximum displacement, when the velocity is v = 0:

gtm + vi = 0

gtm = −vi

tm = −vi/g

where tm is the time to the maximum displacement.

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

Example

If the initial velocity of an object is in the upward direction at 19.6 m/s, what is the velocity at various times?

Solution

vi = −19.6 m/s and g = 9.8 m/s2. Substitute for vi and g in the equation in order to get a formula in terms of t:

v = gt + vi

v = (9.8 m/s2)(t s) + (−19.6 m/s)

Simplify:

v = (9.8t − 19.6) m/s

The table below shows the velocities for different values of t:

t = 0 s v = −19.6 m/s Moving upward from starting point
t = 1 s v = −9.8 m/s Object moving upward
tm = 2 s v = 0 m/s At peak or maximum displacement
t = 3 s v = 9.8 m/s Object moving downward
t = 4 s v = 19.6 m/s Passing starting point
t = 5 s v = 29.4 m/s Continuing downward

Velocities of object projected upward at different times

Velocities of object projected upward at different times

Velocity for displacements moving upward

The general gravity equation for the velocity of an object with respect to the displacement—or movement from the starting point—is:

v = ±√(2gy + vi2)

where

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

Since the velocity is negative (v < 0) on the way up, the negative () version of the equation is used:

v = −√(2gy + vi2)

Also, note that the displacement is negative (y < 0) above the starting point.

Maximum displacement with respect to velocity

At the maximum displacement, the velocity is v = 0. Thus:

−√(2gym + vi2) = 0

2gym + vi2 = 0

2gym = −vi2

ym = −vi2/2g

where ym is the maximum displacement.

If you substitute in values for the displacement where y < ym, the quantity
(2gy + vi2) becomes negative, resulting in the value of √(2gy + vi2) being imaginary or impossible.

Example

If vi = −64 ft/s, find the values of v for various displacements y moving upward.

Solution

Since g = 32 ft/s2, substitute for g and vi in the equation to get a formula in terms of y:

v = −√(2gy + vi2)

v = −√[2*(32 ft/s2)*(y ft) + (−64 ft/s)2]

v = −√(64y ft2/s2 + 4096 ft2/s2)

Since ( ft2/s2) = ft/s, you get:

v = −√(64y + 4096) ft/s

Substitute values for y in the formula, remembering that y is negative above the starting point:

y = 0 ft v = −64 ft/s Moving upward from starting point
y = −32 ft v = −45.3 ft/s Object moving upward
ym = −64 ft v = 0 ft/s At peak or maximum displacement
y = −80 ft -- v = −√(−1024) impossible

Velocity for displacements moving downward

Below the maximum displacement, v is positive, since the object is moving in the direction of gravity. This means that the positive (+) version of the general equation is used:

v = √(2gy + vi2)

However, the displacement is negative from the maximum displacement until the starting point, at which time y = 0. From then on y has positive values.

Example

Continuing the example from above, where vi = −64 ft/s, what are the velocities for various displacements on the way down?

Solution

The positive version of the formula is:

v = √(64y + 4096) ft/s

Substitute values for y in the formula:

ym = −64 ft v = 0 ft/s At peak or maximum displacement
y = −32 ft v = 45.3 ft/s Object moving downward
y = 0 ft v = 64 ft/s Moving downward at starting point
y = 32 ft v = 78.4 ft/s Moving downward below starting point

Velocities for displacements of object projected upward

Velocities for displacements of object projected upward

Summary

When an object is projected upward, it is moving in the opposite direction of the force of gravity, and the initial velocity is a negative number. The velocity is negative while the object moves up and positive while it moves downward. The displacement is negative above the starting point positive below the starting point. The equations for velocity are:

With respect to time

v = gt + vi

With respect to displacement

v = −√(2gy + vi2) going up

vm = 0 (at maximum displacement)

v = √(2gy + vi2) coming down


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