1. Kinematics Free Fall

2. Objectives
State the magnitude and direction of the free fall acceleration on Earth
Explain why all objects fall at the same rate when air resistance is negligible
Sketch the position-time, velocity-time, and acceleration-time graphs for objects moving vertically
Solve free fall problems using the kinematic equations

3. Free Fall Acceleration
Free fall: An object experiencing the acceleration of gravity.
Gravity is a phenomenon associated with mass. Objects have mass (such as the Earth) attract other objects with mass (such as you). The larger and/or closer a mass is, the greater the pull of gravity toward it.
The pull of gravity causes objects with mass to accelerate toward one another. The acceleration of gravity of the Earth has been determined.
Acceleration of gravity of Earth (at or near its surface).
Magnitude: g = 9.8 m/s2
It is often rounded to g = 10 m/s2
Use g = 10 m/s2 for multiple choice, demonstration problems done in class, and if instructed specifically in the text of the problem. Otherwise, use g = 9.8 m/s2
Direction: Down ( −y ) toward the center of Earth.

4. Rewritten in terms displacement (x = x  x0)
Kinematic Equations Adjusted for Vertical Motion
When Used
No final speed
No time
No position
Complete version
Rewritten in terms displacement (x = x  x0)
Object starts at rest
(v0 = 0)
The minus sign before acceleration reflects the direction which is downward and negative. As a result, 9.8 or 10 are substituted as a positive values.
In some problem height, h , and change in height, h , may be used in place of y and y .

5. Kinematic Variables Symbol Variable Positive Negative Zero y0
Initial vertical position
Object is physically located above the origin
Object is physically located below the origin
Object is physically located at the origin y
Final vertical position y
Vertical displacement
Object ends motion lower
Object ends motion higher
Object returns to initial height
v0y
Initial vertical velocity
Initial motion is upward
Initial motion is downward
Dropped from rest vy
Final vertical velocity
Final motion is upward
Final motion is downward
Never g
Acceleration due to gravity
Usually +9.8 is substituted, as equation contains a minus sign

6. Example 1 An object is thrown straight upward at 50 m/s.
a. Determine maximum height the object reaches.
We are only given one value. Initial velocity is 50 m/s.
However, the acceleration of gravity is known, 10 m/s2.
Hidden Zero: Upward moving objects slow until they reach max height. At max height objects are instantaneously at rest, vmax height = 0 m/s. KNOW THIS ! Then they reverse direction and speed up downward.
The object is accelerating and time is not mentioned.
Δy =
v0 =
v =
g =
t =
50 m/s
0 m/s
10 m/s2

7. Example 1 An object is thrown straight upward at 50 m/s.
b. How long does it take to reach maximum height?
We can add the 125 m height at maximum altitude (part “a”) to the list.
To solve for time we can use either:
the displacement-time equation:
which requires the quadratic equation
or the easier velocity-time equation
Δy =
v0 =
v =
g =
t =
125 m
50 m/s
0 m/s
10 m/s2

8. Example 1 An object is thrown straight upward at 50 m/s.
How long does it take to return to it starting point?
The initial launch speed, v0 = 50 m/s, and the acceleration of gravity, g = 10 m/s2 , remain the same.
However, the end point has changed, from max height to landing.
Hidden Zero: Objects thrown upward and returning to their starting point have a vertical displacement of zero, Δy = 0 . KNOW THIS !
We need an equation that addresses displacement and solves for time.
Δy =
v0 =
v =
g =
t =
0 m
50 m/s
10 m/s2

9. Example 1 An object is thrown straight upward at 50 m/s.
d. Determine the final velocity when the object returns to its starting point?
Add the time determined in part “c” to the variable list
We have all the variables except the final velocity, and can now use either velocity equation.
Try the easier equation.
Why −v ? The object was moving downward at 10 s, and down is negative on a coordinate axis.
However, velocity is a vector, and magnitudes of vectors are always positive. Direction can be negative. The answer for velocity should be
Δy =
v0 =
v =
g =
t =
0 m
50 m/s
10 m/s2
10 s

10. Example 1 An object is thrown straight upward at 50 m/s.
e. How does the time to max height compare to the total time of flight?
Time to maximum height: 5 s
Time for entire flight: 10 s
If an object returns to its starting height, then the time to maximum height is half of the total flight time.
f. How does the time moving upward compare to the time moving downward?
They are equal: 5 s up , 5 s down , and 10 s total flight
g. How does the launch velocity compare to the velocity when it returns to the start?
Launch velocity: +50 m/s = 50 m/s, upward
Return velocity: −50 m/s = 50 m/s, downward
They have equal magnitudes (50 m/s), but opposite direction
Note: These facts are only true if the object returns to its original height. If it lands low or high, then the initial and final times and speeds are not equal.

11. An object is thrown downward at 15 m/s from the top of a 150 m tall tower. Determine the objects time of flight.
Example 2
Thrown downward at 15 m/s = 15 m/s
The acceleration of gravity is 10 m/s2
The object begins at the origin (top of the building) and moves 150 m downward, y = 150 m .
You have displacement and need time.
Why was the plus sign selected in the quadratic equation?
For the instant of time that you are solving:
If the object is speeding up, select the plus sign.
If the object is slowing, select the minus sign.
Δy =
v0 =
v =
g =
t =
−150 m
−15 m/s
10 m/s2

12. An object is thrown downward at 15 m/s from the top of a 150 m tall tower. Determine the objects time of flight.
Example 2
There is an alternate method that solves the problem, without using the quadratic equation.
Since time is not known, start with the “no time equation”.
The object is moving downward at the end of the problem, v =  57 m/s
If you use this method it is essential that you consider the direction and sign on the final velocity. If this is wrong, then the answer will be wrong.
Substitute this value into the velocity-time equation.
Δy =
v0 =
v =
g =
t =
−150 m
−15 m/s
10 m/s2

13. An object is dropped from rest from the top of a 45 m tall building.
Example 3
a. Determine the object’s time of flight.
The object is dropped from rest, v0 = 0 m/s.
The object moves downward 45 m, y =  45 m.
The acceleration of gravity is 10 m/s2.
You need the displacement-time equation
This normally requires the quadratic equation. However, initial velocity is zero, and the equation simplifies.
Δy =
v0 =
v =
g =
t =
−45 m
0 m/s
10 m/s2

14. An object is dropped from rest from the top of a 45 m tall building.
Example 3
b. Determine the speed as it impacts the ground.
Add the time from part “a” to the variable list.
Use the velocity time equation.
Since initial velocity is zero, this equation also simplifies
The negative sign on velocity signifies negative direction. The question asks for speed, which is the magnitude of velocity. In kinematics speed is usually reported as a positive value.
v = 30 m/s
Velocity would be reported as: v = 30 m/s, downward.
Δy =
v0 =
v =
g =
t =
−45 m
0 m/s
10 m/s2
0 s

15. Clarification: Objects dropped from rest
When an object is dropped from rest the initial velocity is zero, which results in a simplification of the displacement-time equation.
Dropped objects always move downward, resulting in a displacement that is always negative,  y .
As a result, the equation would become:
And the minus signs will cancel:
Commonly the minus signs are ignored for objects dropped from rest, and the equation is solved using all positive values.
THIS IS ONLY FOR OBJECTS DROPPED FROM REST.