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Explanation of Gravity Time Equations for Objects Projected Downward - Succeed in Understanding Physics. Also refer to physical science, acceleration, velocity, distance, calculation, square-root, Ron Kurtus, School for Champions. Copyright © Restrictions

Gravity Time Equations for Objects Projected Downward

by Ron Kurtus (revised 7 August 2010)

When you throw or project an object downward, it has an initial velocity greater than zero (vi > 0). If you know the initial velocity, there are simple derived equations that allow you to calculate the the time it takes for it to reach a given velocity or when it reaches a given distance from the starting point. Some 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. There is a mini-quiz near the end of the lesson.

Useful tools: Metric-English Conversion | Scientific Calculator.

Time to reach 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 Derivation of Velocity-Time Gravity Equations for details of the derivations.)

Time to reach distance

The equation for the time it takes an object projected downward at some initial velocity to reach a given distance is:

t = [−vi + √(vi2 + 2gy)]/g

where

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

Elapsed time of an object thrown downward as a function of initial velocity and velocity or distance

Elapsed time of an object thrown downward as a function of
initial velocity and velocity or distance

Examples

The following examples illustrate applications of the equations.

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,
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

For a given distance

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,
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 distance from the starting point from the equations:

t = (v − vi)/g

t = [−vi + √(vi2 + 2gy)]/g

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Resources

The following resources provide information on this subject:

Websites

Acceleration due to Gravity Calculations - from Western Washington University

Gravity and Gravitation Resources

Books

Top-rated books on Simple Gravity Science

Top-rated books on Advanced Gravity Physics


Mini-quiz to check your understanding

1. How long does it take to reach 118 m/s if the initial velocity downward is 20 m/s?

1 second

9.8 seconds

10 seconds

2. If t = [−vi + √(vi2 + 2gy)]/g, how long does it take to reach 64 ft if the initial velocity is +32 ft/s?

t = [−32 + √(322 + 2*32*64)]/32 seconds

t = [−32 + √(322 + 2*9.8*64)]/9.8 seconds

t = [32 + √(322 + 2*32*64)]/32 seconds

3. What is the time when the distance is zero?

t = [−vi + √(vi2 )]/g = 2vi/g seconds

t = [−vi + √(vi2 )]/g = 0 seconds

y cannot equal zero

If you got all three correct, you are on your way to becoming a Champion in Physics. If you had problems, you had better look over the material again.


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