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## Gravity Time Equations for Falling Objects

by Ron Kurtus (revised 5 January 2011)

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 time it takes for a falling object to reach a given velocity and the time it takes to reach a given displacement. 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 time to reach a given velocity?
- What is the equation for the time to reach a given displacement?
- What are some examples of these equations?

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

## Time with respect to velocity

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

t = (v − v_{i})/g(

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

Since the initial velocity **v _{i} =** 0 for an object that is simply falling, the equation reduces to:

t = v/g

where

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

Since the object is moving in the direction of gravity, **v** is a positive number.

Elapsed time of a falling object as a function of velocity or displacement

## Time with respect to displacement

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

t = [ −v_{i }± √(v_{i}^{2}+ 2gy) ]/g

where

**±**means plus-or-minus**√(v**is the square root of the quantity_{i}^{2}+ 2gy)**(v**_{i}^{2}+ 2gy)**y**is the vertical displacement in meters (m) or feet (ft)

(

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

When the object is simply dropped, the initial velocity_{} is zero (**v _{i} =** 0) and the equation for elapsed time becomes:

t =_{ }± √(2gy)/g

Since time **t** is always positive, the equation is:

t = √(2gy)/g

Change **g** to **√(g ^{2} )** and simplify the equation:

t =_{ }√(2gy)/√(g^{2})

Thus, the resulting time equation is:

t =_{ }√(2y/g)

## Examples

The following examples illustrate applications of the equations.

### Given the velocity

How long does it take for a falling object to reach 224 ft/s?

#### Solution

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

t = v/g

t= (224 ft/s)/(32 ft/s^{2})

t= 7 seconds

### Given the displacement

How long does it take for an object to fall 200 meters?

#### Solution

Since displacement is in meters, **g** = 9.8 m/s^{2}. Substitute values in the equation:

t = √(2y/g)

t = √[2*(200 m)/(9.8 m/s^{2})]

t = √(40.8 s^{2})

t= 6.39 s

## Summary

There are simple equations for falling objects that allow you to calculate the time taken to achieve a given velocity or displacement. These equations are:

t = v/g

t = √(2y/g)

Check your numbers

## 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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