Derivation of Distance-Time Gravity Equations

The basis for the derivation of the distance-time gravity equations starts with the equation v = gt + v_i that was determined in the Derivation of Velocity-Time Gravity Equations lesson.

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

This lesson will answer those questions. There is a mini-quiz near the end of the lesson.

Basis for distance-time derivations

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

v = gt + v_i

(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)
• v_i 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 distance with respect to time:

v = dy/dt

where

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

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

Derivation of distance with respect to time

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

dy/dt = gt + v_i

Multiply both sides of the equation by dt:

dy = gt*dt + v_i*dt

Integrate dy over the interval from 0 to y:

∫ dy = y

where

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

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

∫gt*dt = gt²/2

Integrate v_i*dt over the interval from 0 to t:

∫v_i*dt = v_it

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

y = gt2/2 + vit

However, the value of the initial velocity affects the equation.

v_i = 0

When the object is simply dropped, the initial velocity is zero (v_i = 0) and the equation becomes:

y = gt²/2

v_i > 0

When the object is thrown downward, the initial velocity is greater than zero
(v_i > 0) and the distance from the starting point is also positive (y > 0).

v_i < 0

When the object is projected upward, it is moving in the opposite direction of the force of gravity, so its initial velocity is less than zero (v_i < 0). Also, the distance y is negative above the starting point—including at the maximum height—and positive below the starting point.

Distance to maximum height

The distance to the maximum or peak height can be derived by starting with the time it takes to reach that height, taken from the Derivation of Velocity-Time Gravity Equations lesson:

t_m = −v_i/g

or

v_i = −gt_m

where t_m is the time to reach the maximum height.

Substitute for v_i in the equation:

y = gt²/2 + v_it

y_m = gt_m²/2 − gt_m²

The resulting equation for the maximum height with respect to the time to reach that height is:

ym = −gtm2/2

Derivation of time with respect to distance

You can find the time it takes for an object to travel a given distance from the starting point by rearranging y = gt²/2 + v_it and solving the quadratic equation for t:

y = gt²/2 + v_it

Subtract y from both sides of the equation and multiply both sides by 2.

gt² + 2v_it − 2y = 0

Solve the quadratic equation for t:

(See Using the Quadratic Equation Formula in our Algebra section for more information.)

Solve the quadratic equation for t. For convenience, we will use the following version of the equation, which is also a more compact form

t = [ −2v_i ± √(4v_i² + 8gy) ]/2g

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

t = [ −2v_i ± 2√(v_i² + 2gy) ]/2g

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

t = [ −vi ± √(vi2 + 2gy) ]/g

where

• ± means plus or minus
• √(v_i² + 2gy) is the square root of the quantity (v_i² + 2yg)

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

Again, the value of the initial velocity affects the equation.

v_i = 0

When the object is simply dropped, the initial velocity is zero (v_i = 0) and the equation becomes:

t = ± √(2gy)/g

Since time t is always positive, the equation is:

t = √(2gy)/g

Change g to √(g² ) and simplify the equation:

t = √(2gy)/√(g² )

t = √(2y/g)

t = √(2y/g)

v_i > 0

When the object is projected downward, v_i and y are positive numbers, and t has only one value. Also, t can only be a positive number. The equation is then:

t = [ −vi + √(vi2 + 2gy) ]/g

v_i < 0

When the object is projected upward, it is moving in a direction opposite of gravity and the initial velocity is less than zero (v_i < 0). However, that means that −v_i is a positive number.

On the way up

Above the starting point, y < 0, and equation used is:

t = [ −vi − √(vi2 + 2gy) ]/g

At maximum height

To find the time it takes to reach the maximum height, rearrange the equation:

y_m = −gt_m²/2

t_m² = −2y_m/g

tm = √(−2ym/g)

Since y_m < 0, −y_m is a positive number.

On the way down

On the way down, y < 0 above the starting point and y > 0 below the starting point. The equation used is:

t = [ −vi + √(vi2 + 2gy) ]/g

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

The derived equations are:

Distance

y = gt²/2 + v_it

y_m = −gt_m²/2

Time

t = [−v_i ± √(v_i² + 2gy)]/g (general equation)

Object dropped

t = √(2y/g)

Projected downward

t = [−v_i + √(v_i² + 2gy)]/g

Projected upward

t = [−v_i − √(v_i² + 2gy)]/g (on way up)

t_m = √(−2y_m/g) (at maximum height)

t = [−v_i + √(v_i² + 2gy)]/g (on way down)

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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

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Mini-quiz to check your understanding

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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