# Derivation of Displacement-Time Gravity Equations

by Ron Kurtus (revised 15 March 2018)

You can derive the gravity equations for the relationship between displacement and time of an object moving in the vertical direction, starting with the derived velocity-time equations.

Note: Displacement is a vector quantity denoting the change in position in a given direction.

Since velocity is the change in displacement over an increment in time, you use Calculus to integrate that change and get the displacement for a given elapsed time. From that displacement equation, you can then determine the equation for the time it takes for the object to reach a given displacement from the starting point.

The derived equations are affected by the initial velocity of the object. This is important in later applications of the equations.

Questions you may have include:

• What is the basis for the derivations?
• What is the displacement for a given time equation?
• What is the time for a given displacement equation?

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

## Basis for displacement-time derivations

To determine the displacement from the starting point for a given time, start with the equation:

v = gt + vi

(Obtained from Derivation of Velocity-Time Gravity Equations)

where

• v is the vertical velocity in m/s or ft/s
• g is the acceleration due to gravity (9.8 m/s2 or 32 ft/s2)
• t is the time in seconds (s)
• vi is the initial vertical velocity in m/s or ft/s

Note: The initial velocity is the velocity at which the object is released after being accelerated from zero velocity. Initial velocity does not occur instantaneously.

Velocity is also the incremental change in displacement with respect to time:

v = dy/dt

where

• dy is the first derivative of vertical displacement y
• dt is the first derivative of time t

By substituting combining these two equations and integrating, you can derive the displacement with respect to time. Then you can rearrange the equation and solve for t to get the time with respect to displacement. Displacement-time relationship

## Derivation of displacement with respect to time

To obtain the displacement with respect to time, substitute for v in v = gt + vi:

dy/dt = gt + vi

Multiply both sides of the equation by dt:

dy = gt*dt + vi*dt

Integrate dy over the interval from y = 0 to y = y:

∫dy = y − 0

where

• is the integral sign between the two limits
• y is the vertical displacement from the starting point

Integrate gt*dt over the interval from t = 0 to t = t:

∫gt*dt = gt2/2 − 0

Integrate vi*dt over the interval from t = 0 to t = t:

∫vi*dt = vit − 0

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

y = gt2/2 + vit

## Derivation of time with respect to displacement

You can find the time it takes for an object to travel a given displacement from the starting point by solving the following quadratic equation for t:

y = gt2/2 + vit

Rearrange the equation by subtracting y from both sides of the equation and multiplying both sides by 2.

gt2 + 2vit − 2y = 0

Solve the quadratic equation for t:

t = [ −2vi ± √(4vi2 + 8gy) ]/2g

Remove the square root of 4 from inside the square root or radical sign:

t = [ −2vi ± 2√(vi2 + 2gy) ]/2g

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

t = [ −vi ± √(vi2 + 2gy) ]/g

where

• ± means plus or minus
• √(vi2 + 2gy) is the square root of the quantity (vi2 + 2yg)

The plus-or-minus sign means that in some situations, there can be two values for t for a given value of y.

## Summary

The basis for the derivation of the displacement-time gravity equations starts with the equation v = gt + vi. Since velocity is the change in displacement over an increment in time, you integrate that change and get the displacement for a given elapsed time.

From that displacement equation, you can then determine the equation for the time it takes for the object to reach a given displacement from the starting point.

The derived equations are:

y = gt2/2 + vit

t = [−vi ± √(vi2 + 2gy)]/g

Find solutions to your problems by being clever

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