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Gravity Time Equations for Objects Projected Downward

by Ron Kurtus (revised 7 January 2011)

When you throw or project an object downward, it is accelerated until it is released at some initial velocity. If you know this initial velocity, there are simple derived equations that allow you to calculate the time it takes for it to reach a given velocity or when it reaches a given displacement from the starting point.

Examples illustrate these equations.

Note: You normally do not need to memorize these equations, but you should know where to find them in order to solve equations.

Questions you may have include:

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



Time with respect to velocity

The equation for the time it takes an object that is thrown or projected downward to reach a given velocity is:

t = (v − vi)/g

where

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

Since the object is moving in the direction of gravity, v and vi are positive numbers.

Time with respect to displacement

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

t = [−vi ± √(vi2 + 2gy)]/g

where

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

Since vi is downward, it has a positive value and −vi is obviously negative. This means that the + version of the equation must be used in order to make t a positive number. The equation is then:

t = [−vi + √(vi2 + 2gy)]/g

Since y is below the starting point, it also is a positive number.

Time as a function of downward velocity or displacement

Time as a function of downward velocity or displacement

Examples

The following examples illustrate applications of the equations.

Time for a given velocity

If you throw a ball downward from a tall building at 5 ft/s, find the time it takes for the ball to reach a velocity of 101 ft/s.

Solution

You are given that vi = +5 ft/s and v = 101 ft/s. Since vi and v are in ft/s, then
g = 32 ft/s2. The equation to use is:

t = (v − vi)/g

Substitute values in the equation:

t = (101 ft/s − 5 ft/s)/(32 ft/s2)

t = (96 ft/s)/(32 ft/s2)

t = 3 s

Time for a given displacement

If you throw an object downward from a high building at 5 m/s, find the time it takes to fall 50 m.

Solution

You are given that vi = +5 m/s and y = 50 m. Since vi in m/s and y is in m, then
g = 9.8 m/s2. The equation to use is:

t = [−vi + √(vi2 + 2gy)]/g

Substitute values in the equation:

t = [−5 m/s + {(25 m/s)² + 2*(9.8 m/s²)*(50 m)}]/(9.8 m/s²)

t = [−5 m/s + (625 m²/s² + 980 m²/s²)]/(9.8 m/s²)

t = [−5 m/s + (1605 m²/s²)]/(9.8 m/s²)

t = [−5 m/s + 40.1 m/s]/(9.8 m/s²)

t = (35.1 m/s)/(9.8 m/s²)

t = 3.58 s

(Whew!)

Summary

You can calculate the time it takes an object that is projected downward to reach a given velocity or reach a given displacement from the starting point from the equations:

t = (v − vi)/g

t = [−vi + √(vi2 + 2gy)]/g


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Resources and references

Ron Kurtus' Credentials

Websites

Gravity Resources

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

Do you have any questions, comments, or opinions on this subject? If so, send an email with your feedback. I will try to get back to you as soon as possible.


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