Freely Falling Objects
Freely Falling Objects
• Important & common special case of
uniformly accelerated motion:
“FREE FALL”
Objects falling in Earth’s gravity.
Neglect air resistance. Use one dimensional
uniform acceleration equations (with some
changes in notation, as we will see)
Near the surface of the Earth, all objects experience
approximately the same acceleration due to gravity.
This is one of the most
common examples of motion
with constant acceleration.
In the absence of air
resistance, all objects fall
with the same acceleration,
although this may be tricky to
tell by testing in an
environment where there is
air resistance.
Falling Objects
• Experiment:
– Ball & light piece of paper dropped at the
same time. Repeated with wadded up paper.
• Experiment:
– Rock & feather dropped at the same time in air.
Repeated in vacuum.
The acceleration due to gravity
at the Earth’s surface is
Approximately 9.80 m/s2.
At a given location on the
Earth & in the absence of
air resistance, all objects
fall with the same
constant acceleration.
• Experiment finds that the acceleration of
falling objects (neglecting air resistance) is
always (approximately) the same, no
matter how light or heavy the object.
• The magnitude of the acceleration
due to gravity,
ag
g = 9.8 m/s2 (approximately!)
• The acceleration of falling objects is always
the same, no matter how light or heavy.
• Acceleration due to gravity, g = 9.8 m/s2
• First proven by Galileo Galilei
A Legend:
He dropped objects
off of the leaning
tower of Pisa.
• The magnitude of the acceleration due to
gravity: g = 9.8 m/s2
(approximately!)
• It has a slight dependence on the location on Earth,
on the latitude & the altitude:
A Common Misconception!
The Algebraic Sign (+ or - ?) in the
Kinematics Equations of the
One-Dimensional Vector
Gravitational Acceleration
• Note: My treatment is slightly different than the book’s, but it is equivalent!
• To treat motion of falling objects, we use the same
equations we already have, but change notation slightly:
Replace a by g = 9.8 m
BUT in the equations, it could have
a + or a - sign in front of it!
We discuss this next!
• Usually, we consider vertical motion to be in the y
direction & so replace x by y and x0 by y0 (often y0 = 0)
NOTE!!!
Whenever I (or the author!) write the
symbol g, it ALWAYS means the
POSITIVE numerical value 9.8 m/s2!!
It is NEVER negative!!! The sign (+ or
-) of the one-dimensional gravitational
acceleration VECTOR is taken into
account in the Equations we now
discuss!
The Sign of g in the 1d Equations
• The magnitude (size) of g = 9.8 m/s2 (POSITIVE!)
– But, acceleration is a (1 dimensional)
VECTOR with 2 possible directions.
– Call these + and -.
– However, which way is + and which way is – is
ARBITRARY & UP TO US!
– It may seem “natural” for “up” to be + y & “down” to
be - y, but we could also choose (we sometimes will!)
“down” to be + y and “up” to be - y
So, in the equations g could have a + or a sign in front of it, depending on our choice!
Directions of Velocity & Acceleration
• Objects in free fall
ALWAYS have DOWNWARD acceleration.
• We still use the same equations for objects thrown
upward with some initial velocity v0
• An object goes up until it stops at some point & then it
falls back down. The acceleration vector is always g in
the downward direction. For the first half of flight, the
velocity is UPWARD.
 For the first part of the flight, velocity &
acceleration are in opposite directions!
VELOCITY & ACCELERATION
ARE NOT NECESSARILY IN
THE SAME DIRECTION!
Equations for Objects in Free Fall
• Written taking “up” as + y
v = v0 - gt
(1)
y = y0 + v0t - (½)gt2
(2)
(v)2 = (v0)2 - 2g(y - y0)
(3)
vavg = (½)(v + v0)
(4)
g = 9.8 m/s2
Often, y0 = 0. Sometimes v0 = 0
Equations for Objects in Free Fall
• Written taking “down” as + y
v = v0 + gt
(1)
y = y0 + v0t + (½)gt2
(2)
(v)2 = (v0)2 + 2g(y - y0)
(3)
vavg = (½)(v + v0)
(4)
g = 9.8 m/s2
Often, y0 = 0. Sometimes v0 = 0
Free Fall Examples
Example
Falling from a tower (v0 =
0)
Note!
Take y as positive
DOWNWARD!
v = at
y = (½)at2
a = g = 9.8 m/s2
Example
Falling from a tower (v0 =
0)
Note!
Take y as positive
DOWNWARD!
v = at
y = (½)at2
a = g = 9.8 m/s2
Example
Falling from a tower (v0 =
0)
Note!
Take y as positive
DOWNWARD!
v = at
y = (½)at2
a = g = 9.8 m/s2
Example
v1 = (9.8)(1) Falling from a tower (v0 = 0)
= 9.8 m/s
v2 = (9.8)(2)
= 19.6 m/s
v3 = (9.8)(3)
= 29.4 m/s
Note!
Take y as positive
DOWNWARD!
v = at
y = (½)at2
a = g = 9.8 m/s2
Example:
Thrown Down From a Tower
A ball is thrown downward with initial
velocity v0 = 3.0 m/s, instead of being
dropped.
(a) Position after t = 1.0 s & 2.0 s?
y = v0t + (½)at2
t = 1.0 s; y = (3)(1) + (½)(9.8)(1)2 = 7.9m
t = 2.0s ; y = (3)(2) + (½)(9.8)(2)2 = 25.6 m
Photo of the leaning
tower of Pisa
Example:
Thrown Down From a Tower
A ball is thrown downward with initial
velocity v0 = 3.0 m/s, instead of being
dropped.
(a) Position after t = 1.0 s & 2.0 s?
y = v0t + (½)at2
t = 1.0 s; y = (3)(1) + (½)(9.8)(1)2 = 7.9m
t = 2.0s ; y = (3)(2) + (½)(9.8)(2)2 = 25.6 m
Photo of the leaning
tower of Pisa
Example:
Thrown Down From a Tower
A ball is thrown downward with initial
velocity v0 = 3.0 m/s, instead of being
dropped.
(a) Position after t = 1.0 s & 2.0 s?
y = v0t + (½)at2
t = 1.0 s; y = (3)(1) + (½)(9.8)(1)2 = 7.9m
t = 2.0s ; y = (3)(2) + (½)(9.8)(2)2 = 25.6 m
(b) Speed after t = 1.0 s & 2.0 s?
v = v0 + at
t = 1.0 s; v = 3 + (9.8)(1) = 12.8m/s
t = 2.0s ; v = 3 + (9.8)(2) = 22.6m/s
Photo of the leaning
tower of Pisa
Compare with speeds of a dropped ball.
A Useful, Detailed Example
v = 0 here
but still have
a = -g
A person throws a ball up into the
air with initial velocity v0 = 15.0 m/s.
Time to the top
= (½) round trip
time!
a. Time to the top?
b. Round trip time?
c. Maximum height?
d. Velocity when it comes back
to the start?
e. Times when the height y = 8.0 m?
Questions:
vC = -vA
(= -v0)
Work in general on
the white board!
vA = v0 = 15 m/s
choose y as positive upward
 a = -g = - 9.8 m/s2
Example: Ball thrown upward; the quadratic formula.
For a ball thrown upward at an initial speed of v0 =
15.0 m/s, calculate the times t the ball passes a point y
= 8.0 m above the person’s hand.
Example: A ball thrown upward at the edge of a cliff.
A ball is thrown up with initial velocity
v0 = 15.0 m/s, by a person standing on
the edge of a cliff, so that it can fall to
the base of the cliff 50.0 m below.
Calculate:
a. The time it takes the ball to
reach the base of the cliff.
b. The total distance traveled by
the ball.
Example: Not a bad throw for a rookie!
A stone is thrown at point (A) from the top
of a building with initial velocity v0 = 19.2
m/s straight up. The building is H = 49.8 m
high, & the stone just misses the edge of the
roof on its way down, as in the figure.
Calculate:
a) The time at which it reaches its
maximum height.
b) It’s maximum height above the rooftop.
c) The time at which it returns to the
thrower’s hand.
d) It’s velocity when it returns to the
thrower’s hand.
e) It’s velocity & position at time t = 5 s.