One dimensional motion
2.3 free fall
Objectives
1. Relate the motion of a freely falling body to motion with
constant acceleration.
2. Calculate displacement, velocity, and time at various
points in the motion of a freely falling object.
3. Compare the motion of different objects in free fall.
Homework due: 10/4
Test: 10/4
Key words and
ideas
Free fall
Notes
Text Page 60 Figure 2.14
• If the feather and apple experiment were
performed on the moon, where free-fall
acceleration is approximately on-sixth the
value of free-fall acceleration on Earth,
how would the picture compare with figure
2-14?
Text Page 61 Figure 2.15
• If the initial velocity of the ball is 10.5 m/s
when it is thrown from the ground, what is
the final velocity of the ball at the end of its
flight?
Text Page 61 Figure 2.16
• How do you describe the motion from this
graph?
• At what time is the velocity zero?
• At what time is the acceleration zero?
Two important motion characteristics that
are true of free-falling objects
• Free-falling objects do not
encounter air resistance.
• All free-falling objects (on Earth)
accelerate downwards at a
constant rate.
“g” - The “Magic” Number
• “g” means ACCELERATION DUE TO GRAVITY
• Each planet, star, moon, or other large object
has its own value for “g”
Examples
• “g” is 1.62 m/s2 on the Moon
• “g” is 26 m/s2 on Jupiter
• “g” is 9.81 m/s2 on Earth
If the feather and elephant experiment were performed on
the moon, would they still fall at the same rate, just like
on Earth?
a = -g = -9.81m/s2 for all free fall objects.
(the ‘-’ sign means “downward”)
•
•
•
•
Dropped from rest
Throwing downward
Throwing upward
Throwing side ways
• Since free fall motion has constant
acceleration, we can apply all
kinematics equations on free fall
motion.
Free fall – motion with constant acceleration
a = -g
To solve free fall motion problems, we can use kinematics
equations with constant acceleration.
a = -g
∆x = ∆y
Velocity of a dropped object
• An object falls from rest. What is its
velocity at the end of one second? Two
= -9.81seconds?
m/s2 (‘-’ means “downward”)
v f  viseconds?
 at a = -gThree
t=0
vi = 0 m/s
t=1s
v = -9.81 m/s
t=2s
v = -19.62 m/s
t=3s
v = -29.43 m/s
Displacement of a dropped object
• An object falls from rest. How far has it
fallen at the end of one1 second? Two
d seconds?
vi t  at 2
seconds? Three
2
t=0
d=0m
t=1s
d = -4.905 m
t=2s
d = -19.62 m
t=3s
d = -44.145 m
Velocity and distance of a free falling object
dropped from rest
+
Throwing Downward
av=i =-9.81
-10 m/s 2
• An object is thrown downward from
the top of a 175 meter building with
an initial speed of 10 m/s.
– What acceleration does it
experience?
• -9.81 m/s2
– What is the object’s initial
velocity?
• -10 m/s
“a” and “vi” in SAME DIRECTION
Building Example (cont.)
• What is the object’s velocity as it hits the
ground?
•
•
•
•
vi = -10 m/s
d = -175 m
a = -9.81 m/s2
vf = ?
vf2 = vi2 + 2ad
vf2 = (10m/s)2 + 2(-9.81 m/s2)(-175 m)
vf = -59.4 m/s
• How long does it take the object to hit the
vf = vi + at
ground?
-59.4 m/s = -10 m/s + (-9.81 m/s2) t
• t=?
t = 5.05 s
Throwing Upward
t=2s
v = +30.38 m/s
t=1s
v = +40.19 m/s
t=0s
vi = +50 m/s
v f  vi  at
• A cannon fires a shot
directly upward with an
initial velocity of 50 m/s.
– What acceleration does
the cannonball
experience?
• -9.81 m/s2
– What is the cannonball’s
initial velocity?
• +50 m/s
“a” and “vi” in OPPOSITE
DIRECTION
Cannon (cont.)
• What is the object’s velocity as it reaches the
top of its flight?
0 m/s
(All objects momentarily STOP at the top of their flight)
• How long does it take the cannonball to reach
the top of its flight?
vf = vi + at
0 = +50 m/s + (-9.81 m/s2) t
t = 5.1 s
• What is the maximum height of the
cannonball?
d = v t + ½ at2
i
d = (+50 m/s)(5.1 s) + ½ (-9.81 m/s2)(5.1 s)2
d = +127.4 m
• A rocket is fired straight up into the air to
an altitude of 20 m.
1.At the top of its flight, what is its velocity? .
2.At the top of its flight, what is its
acceleration?
3.After the rocket falls back to its launching
pad, what is its displacement?
At the top of an object’s flight, its velocity is zero. Why its
acceleration is not zero?
Page 63 - Sample problem 2F
• Jason hits a volleyball so that it moves with an
initial velocity of 6.0 m/s straight upward. If the
volleyball starts from 20 m above the floor, how
long will it be in the air before it strikes the floor?
Assume that Jason is the last player to touch the
ball before it hits the floor.
Class work
• Page 64 – practice 2F
Free Fall Graphs of a dropped object
A position versus time graph
for a free-falling object
Disp.
velocity
velocity increase in the
negative direction – slope
is neg. & increasing
A velocity versus time graph
for a free-falling object
velocity increases in the
negative direction at
constant rate – slope is neg.
& constant
t
t
Free Fall Graphs of upward object
velocity
Upward: velocity is big, positive, decreasing,
slope is constant (a = -9.8 m/s/s).
Top, velocity is zero. Slope remains the same
(acceleration is still -9.8 m/s/s)
position
time
Downward: velocity increases in negative
direction at the same constant rate of -9.81
m/s/s (slope remains the same), reaches the
same speed as it started upward.
Upward: displacement increases, slope is
positive, decreasing (velocity is positive,
decreasing)
Top: slope = 0 (its velocity is zero)
time
Downward: displacement decreases, its slope
increases in negative direction (velocity is
negative, increasing)
Review
1. An object accelerates uniformly from rest to a speed of
50. m/s in 5.0 seconds. What is the average speed of
the object during the 5.0 second interval?
2. A car travels 90. meters due north in 15 seconds. Then
the car turns around and travels 40. meters due south
in 5.0 seconds. What is the magnitude of the average
velocity of the car during this 20.-second interval?
3. A locomotive starts at rest and accelerates at 0.12
meters per second squared to a speed of 2.4 meters
per second in 20. seconds. Describe its velocity and
acceleration using words “constant” and “increasing”.
4. A race car traveling at 10. m/s accelerates at a rate of
1.5 m/s2 while traveling a distance of 600.
meters. What is the final speed of the race car?
do now
• What is free fall? (2 conditions)
Which pair of graphs represents the same motion?
A.
B.
C.
D.
• A cart travels with a constant nonzero acceleration along
a straight line. Which graph best represents the
relationship between the distance the cart travels and
time of travel?
A.
C.
B.
D.
• A car traveling at a speed of 13 meters per
second accelerates uniformly to a speed
of 25 meters per second in 5.0 seconds.
• A truck traveling at a constant speed
covers the same total distance as the car
in the same 5.0- second time interval.
Determine the speed of the truck.
• Answer: ______________ m/s
•
A.
B.
C.
D.
A skier starting from rest skis straight
down a slope 50. meters long in 5.0
seconds. What is the magnitude of the
acceleration of the skier?
20. m/s2
9.8m/s2
5.0 m/s2
4.0 m/s2
•
1.
2.
3.
4.
When velocity is zero and acceleration is
negative, what happens to the object’s
motion?
The object slows down.
The object speeds up from a negative
velocity.
Nothing happens to the object.
The object speeds up from rest.
10/2 do now
•
1.
2.
3.
4.
A trucker drives with an average velocity
of 27 m/s toward the east on a perfectly
straight highway. What will the trucker’s
displacement be after 2.0 h?
97 km to the east
1.6 × 102 km to the east
1.9 × 102 km to the east
3.2 × 102 km to the east
• Frame of reference
– Toward each other – add speed
– Move in the same direction – subtract speed
• Vector, scalar
– Vector has magnitude (number) and direction
– Scalar has magnitude only
• Distance, displacement (x, y, d, ∆x, ∆y)
– Distance – scalar, length of the path
– Displacement – vector, direct distance
v = d/t; v = ½ (vi + vf)
aavg = ∆v/∆t = (vf - vi)/∆t
vf = vi + a∆t
d = vit + ½ at2
vf2 = vi2 + 2ad
• Speed, velocity, average velocity, instantaneous velocity (v, v)
– vavg = ∆x/∆t; vavg = d/t; Unit: m/s
d
d
– v = slope of d vs. t graph;
t
t
• Acceleration (a)
– aavg = ∆v/∆t = (vf-vi)/∆t, unit: m/s2
– aavg = slope of v vs. t graph;
– Disp. = area under v vs. t graph;
v
v
t
t
• Free fall - two conditions: (g = 9.81 m/s2)
–
–
–
–
No air resistance, all objects falls the same distance during same time
a = - g;
v
d
tup = tdown; vup = -vdown;
t
t
at top: v = 0, a = -g
Lab 5 – time interval of free fall
• Objective: determine your reaction time
• Material: ruler
• Data table – show work for one trial:
trial
1
2
3
∆y (m)
Do this lab with the
person sits next to
you
4
5
Average
• conclusion: use your data to determine your reaction
time: use equation, substitute number and units and
answer with units.
Do now
• A trucker drives with an average velocity
of 27 m/s toward the east on a perfectly
straight highway. What will the trucker’s
displacement be after 2.0 h?
a. 97 km to the east
b. 1.9  102 km to the east
c. 1.6  102 km to the east
d. 3.2  102 km to the east
Lab 3 – Position vs. Time graph (5 pt.)
• Purpose: determine the position-time graph of an electric
car moving at constant velocity (5 pt)
• Material: electric car, ticker tape, spark timer, ruler (5 pt)
• Procedure: briefly describe how you measured your data so
that others will know to follow your steps to complete this lab
(10 pt)
Time (s)
Distance (cm)
• Data table: (20 pt)
0
0
• Data Analysis: graph data;
determine slope – must show
work; what is the meaning of
your slope? (30 pt)
0.1
0.2
0.3
0.4
0.5
Once you finish your lab, you may correct lab 2 and do
problems in your guided notes – page 11 & 12