Gravity Time Equations for Objects Projected Downward
About 3 min read
(updated 29 May 2023)
When you throw or project an object downward, it is accelerated until it is released at some 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:
- How do you find the time for a given velocity?
- How do you find the time for 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 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
- t is the time in seconds (s)
- v is the vertical velocity of the falling object in feet/second (ft/s) or meters/second (m/s)
- vi is the initial vertical velocity the object has been projected downward in ft/s or m/s
- g is the acceleration due to gravity; g = 32 ft/s2 or 9.8 m/s2
Time with respect to displacement
The general gravity equation for the time with respect to displacement is:t = [−vi ± √(vi2 + 2gy)]/g
where
- ± means plus or minus
- y is the vertical displacement in feet (ft) or meters (m)
- √(vi2+ 2gy) is the square root of the quantity (vi2+ 2gy)
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
Gravitate toward good grades
Resources and references
Websites
Equations for a falling body – Wikipedia
Gravity Calculations – Earth – Calculator
Kinematic Equations and Free Fall – Physics Classroom