An object experiencing the acceleration of gravity is described. The acceleration of gravity on Earth is approximately 9.8 m/s2 downward. Equations of motion are provided for falling bodies, including simplified equations that assume the object starts from rest. Examples are worked through comparing the traditional method of using negative signs for downward motion to an alternative method where downward is considered positive to simplify the signs.

1. KKiinneemmaattiiccss 11DD
FFrreeeeffaallll
© RHJansen

2. FFaalllliinngg BBooddyy
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.
© RHJansen

3. Adapting tthhee KKiinneemmaattiicc EEqquuaattiioonnss
1
2
© RHJansen
Kinematic Equations
Modified for y direction
Kinematic Equations
Simplified using Δ
Simplified if objects are
dropped from rest v0 = 0
1
2
y=y0+v0t+
gt2
v=v0+gt
2+2gy-y0 ( )
v2=v0
Dy=v0t+
gt2
v=v0+gt
v2=v0
2+2gDy
Dy=
1
2
gt2
v=gt
v= 2gDy
1st: The easiest problems are those starting at rest.
If the object starts at rest use the three equations in the right column
2nd: Another easy problem is when time is given.
If I have time I will use the top two equations
3rd: The most difficult problems do not give time.
In these situations the equation missing time is the best starting point.

4. ZZeerroo’’ss,, PPlluusssseess,, aanndd MMiinnuusseess
These are often the trick to the problem and hidden in the text.
Zero Positives and Negatives
y0 Start at origin + Starts above origin
− Starts below origin
y Ends at origin + Ends above origin
− Ends below origin
© RHJansen
Δy At maximum altitude
(maximum height)
+ Final position is above initial position
− Final position is below initial position
v0 Initially at rest + Moving upward
− Moving downward
v Stops + Moving upward
− Moving downward
g
Can’t be zero. Always
gravity due to all the
mass in the universe. Can
be negligible if deep in
space.
− Moving upward (+) and slowing down (−)
− Moving downward (−) and speeding up (+)
We will learn a trick solution setting gravity as
positive, even though it is truly always negative.
t N/A + Always

5. EExxaammppllee 11
An object is thrown straight upward at 50 m/s.
© RHJansen
a. Determine its maximum altitude.
Seems like we have very little information.
•We are only given one value. Initial velocity is 50 m/s.
•However, in gravity problems the acceleration of gravity is known, −10 m/s2.
(I use −10 m/s2 in class to make solving easier. For homework use −9.8 m/s2)
•The key to this problem is an important hidden zero. At maximum altitude
(height) objects come to an instantaneous stop and reverse direction. When a
problem ends at maximum height velocity final is zero, v = 0 m/s .
This will happen throughout the year. Know it ! Know it ! Know it !
50 m/s
0 m/s
v2=v0
2+2gDy
Δy =
v0 =
v =
g =
t =
−10 m/s2
This problem is accelerating and time is not mentioned.
Dy =
v2 - v0
2
2g
=
(0)2 - (50)2
2(-10) = 125m

6. EExxaammppllee 11
An object is thrown straight upward at 50 m/s.
b. How long does it take to reach maximum altitude?
We can add the 125 m height at maximum altitude (part “a”) to the list.
125 m We are looking for time and have all other data.
We can use either of the equations containing time.
•However, the displacement equation
would require the use of the quadratic equation.
•On the other hand the velocity equation is simple.
v = v0 + gt
© RHJansen
Δy =
v0 =
v =
g =
t =
50 m/s
0 m/s
−10 m/s2
t =
v - v0
g
=
(0)- (50)
(-10) = 5s
Dy = v0t + 1
2
gt2

7. EExxaammppllee 11
An object is thrown straight upward at 50 m/s.
c. How long does it take to return to the ground?
The end point has changed. This is a new problem with a new variable list.
0 m It still has the same launch speed or 50 m/s.
Gravity also remains the same at −10 m/s2.
At maximum height the speed was known. However,
when an object thrown upward returns to the ground
the height is displacement is known, Δy = 0 .
Accelerating and time is missing.
© RHJansen
v = v0 + gt
Δy =
v0 =
v =
g =
t =
50 m/s
−10 m/s2
t =
v - v0
g
v2 = v0
2 + 2gDy
2 + 2gDy
v = ± v0
v = ± (50)2 + 2(-10)(0)
v = -50m s
Why −v ?
The object
was returning
to the ground.
It was moving
downward,
and down is
negative.
t =
(-50)- (50)
(-10) = 10s

8. EExxaammppllee 11
An object is thrown straight upward at 50 m/s.
d. How does the time to maximum height compare to the total time?
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.
e. How does the time going up compare to the time going down?
They must be equal: 5 s up , 5 s down , and 10 s total flight
f. How does the launch velocity compare to the velocity when it returns?
Launch velocity +50 m/s
Velocity when object returns to starting height: −50 m/s
They have equal magnitude (50 m/s), but opposite direction
(+ moving up / − moving down)
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.
© RHJansen

9. TThhee PPrroobblleemm WWiitthh AAcccceelleerraattiioonn
Acceleration can have two signs
1. Acceleration is a vector quantity with direction along an axis
denoted with plus or minus.
Accelerating in the +x or +y direction: +a
Accelerating in the −x or −y direction: −a
2. Acceleration is the rate of change in velocity
When velocity increases ( v > v0 ): +a
When velocity decreases ( v < v0 ): −a
+a could be moving right or up ( + ) and increasing in speed ( + ), or it
could be moving left or down ( − ) and decreasing in speed ( − ).
−a could be moving right or up ( + ) and decreasing in speed ( − ), or it
could be moving left or down ( − ) and increasing in speed ( + ).
© RHJansen
a =
v - v0
t
a = Dv
t

10. There is aa ttrriicckk ttoo ssiimmpplliiffyyiinngg tthhee ssiiggnnss
Eliminate one of the signs: Eliminate the sign on direction.
Simply always make direction Positive. How?
Always set the objects initial direction of motion as the positive direction.
Example for gravity problems:
Object is thrown upward: Set upward as positive.
Object is thrown downward: Set downward as positive.
Benefit of this trick
Initial velocity is always positive: +v0
If an object is speeding up initially: +a
If an object is slowing down initially: −a
The direction you declare as positive does not matter. It is important that
your decision remain in force for the entire problem. In addition, the signs
you place on all other quantities must match your decision.
© RHJansen

11. There is aa ttrriicckk ttoo ssiimmpplliiffyyiinngg tthhee ssiiggnnss
Consistency is key
Example: Object moves down initially: Set down as positive.
The coordinate axis is now upside down for the entire problem.
View the problem as though you are standing on your head.
Or, turn the page upside down if you have to.
Since down is positive all the other signs must match
+Δy If thrown down it is displaced down, and down is set as
positive.
+vThis determined the positive direction. Sets the axis for this
0 trick.
+v At the end it will still be moving down, and down is set as
positive.
+g Acceleration always points down, and down is set as positive.
Does this trick always work?
No ! Two problems that demand the conventional coordinate system.
Graphing: right and © up RHJansen
are always positive. Do not use the
trick.

12. EExxaammppllee 22
An object is thrown downward at 15 m/s from the top of a
150 m tall tower. Determine the objects time of flight.
This will be solved using both the traditional method and using the trick.
Traditional: −150 m Δy =
+150 m
Up is always
−15 m/s
positive
+15 m/s
2 + 2gDy
−57 m/s +57 m/s
2 + 2gDy
Down is negative: −v answer Down is positive: +v answer
© RHJansen
Δy =
v0 =
v =
g =
t =
−10 m/s2
v0 =
v =
g =
t =
+10 m/s2
Trick Method:
v0 is down
Set down positive
Accelerating,
and time is
unknown.
Accelerating,
and time is
unknown.
v2 = v0
2 + 2gDy
v = ± v0
v = ± (-15)2 + 2(-10)(-150)
v2 = v0
2 + 2gDy
v = ± v0
v = ± (+15)2 + 2(+10)(+150)
v = -57m s v = +57m s
Add to variable list and continue. Add to variable list and continue.

13. EExxaammppllee 22
An object is thrown downward at 15 m/s from the top of a
150 m tall tower. Determine the objects time of flight.
This will be solved using both the traditional method and using the trick.
Traditional: −150 m Δy =
+150 m
Up is always
−15 m/s
positive
−57 m/s +57 m/s
© RHJansen
Δy =
v0 =
v =
g =
t =
−10 m/s2
v0 =
v =
g =
t =
+15 m/s
+10 m/s2
Trick Method:
v0 is down
Set down positive
Have all
variables. Use
easy equation.
Have all
variables. Use
easy equation.
t =
v - v0
g
Always works, but involves a lot of
negative signs. They cancel, but
forgetting even one creates problems.
t =
v - v0
g
Eliminates all negative signs for an
easier solution. Ideal for motion in
one direction that does not reverse.
v = v0 + gt
t = 4.2s
=
(-57)- (-15)
(-10)
v = v0 + gt
t = 4.2s
=
(+57)- (+15)
(+10)

14. EExxaammppllee 33
An object is dropped from rest from the top of a 45 m tall
building
a. Determine the speed as it impacts the ground.
Traditional: −45 m Δy =
+45 m
Up is always
0 m/s
positive
0 m/s
−30 m/s +30 m/s
Down is negative: −v answer Down is positive: +v answer
© RHJansen
Δy =
v0 =
v =
g =
t =
−10 m/s2
v0 =
v =
g =
t =
+10 m/s2
v0 sets direction,
but v0 = 0 .
When released it
does move down
initially, so set
down as positive.
v2 = v0
2 + 2gDy
v = ± 2gDy
v = ± 2(-10)(-45)
v2 = v0
2 + 2gDy
v = ± 2gDy
v = ± 2(+10)(+45)
v = -30m s v = +30m s
Add to variable list and continue. Add to variable list and continue.

15. EExxaammppllee 33
An object is dropped from rest from the top of a 45 m tall
building
Traditional: −45 m Δy =
+45 m
Up is always
0 m/s
positive
© RHJansen
b. Determine the time of flight.
Δy =
v0 =
v =
g =
t =
−10 m/s2
v0 =
v =
g =
t =
0 m/s
+10 m/s2
v0 sets direction,
but v0 = 0 .
When released it
does move down
initially, so set
down as positive.
−30 m/s +30 m/s
v = v0 + gt
t =
v - v0
g
t = 3.0s
=
(-30)- (0)
(-10)
v = v0 + gt
t =
v - v0
g
t = 3.0s
=
(+30)- (0)
(+10)