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# Gravity Displacement Equations for Falling Objects

by Ron Kurtus (revised 9 Jun 2011)

The displacement of an object is the change in position from the starting point in a specific direction and can be represented as a vector. It is different from distance, where direction is not indicated.

When you drop an object from some height above the ground, it has an initial velocity of zero. Simple equations allow you to calculate the * displacement the object falls* until it reaches a given velocity or after a given period of time. The equations assume that air resistance is negligible.

Examples demonstrate applications of the equations.

Questions you may have include:

- What is the equation for the displacement to reach a given velocity?
- What is the equation for the displacement for a given time?
- What are some examples of these equations?

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

## Displacement with respect to velocity

The general gravity equation for displacement with respect to velocity is:

y = (v^{2}− v_{i}^{2})/2g(

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

Since the initial velocity **v _{i} =** 0 for a dropped object, the equation reduces to:

y = v^{2}/2g

where

**y**is the vertical displacement in meters (m) or feet (ft)**v**is the vertical velocity in m/s or ft/s**g**is the acceleration due to gravity (9.8 m/s^{2}or 32 ft/s^{2})

Since the object is moving downward from the starting point, both **y** and **v** are positive numbers.

Displacement of a falling object as a function of velocity or time

## Displacement with respect to time

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

y = gt^{2}/2 + v_{i}t(

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

Since **v _{i} =** 0 for a dropped object, the equation reduces to:

y = gt^{2}/2

where **t** is the time in seconds (s).

## Examples

The following examples illustrate applications of the equations.

### Given the velocity

If **v** = 75 ft/s, how far has the object fallen?

#### Solution

Since **v** is in ft/s, **g** = 32 ft/s^{2}. Substitute values in the equation:

y = v^{2}/2g

y= (75 ft/s)*(75 ft/s)/2*(32 ft/s^{2})

y= (5625 ft^{2}/s^{2})/(64 ft/s^{2})

y= 87.89 ft

### Given the elapsed time

If **t** = 4 seconds and **g** = 9.8 m/s^{2}, find the displacement the object has fallen.

#### Solution

Substitute values in the equation:

y = gt^{2}/2

y= (9.8 m/s^{2})*(16 s^{2})/2

y= 78.4 m

## Summary

Displacement is the change in position from the starting point in a specific direction. The following equations allow you to calculate the displacement the object falls until it reaches a given velocity or after a given period of time:

y = v^{2}/2g

y = gt^{2}/2

Have confidence in yourself

## Resources and references

### Websites

**Falling Bodies** - Physics Hypertextbook

**Equations for a falling body** - Wikipedia

**Gravity Calculations - Earth** - Calculator

**Kinematic Equations and Free Fall** - Physics Classroom

### Books

**Top-rated books on Simple Gravity Science**

**Top-rated books on Advanced Gravity Physics**

## Questions and comments

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## Gravity Displacement Equations for Falling Objects