Falling Objects
Freely Falling Objects
• An important & common special case of
uniformly accelerated motion is
“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)
Falling Objects
Experiment
• Ball & light piece of paper dropped at the same
time. Repeated with wadded up paper.
Experiment
• A rock & a feather are dropped at the same
time in air. Repeated in vacuum.
• 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.
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
Acceleration Due to Gravity
g = 9.8 m/s2
(approximately)
• Depends on location on Earth, latitude, & altitude:
• Note: My treatment is slightly different than the
book’s, but it is, of course, equivalent!
• To treat falling objects, use the same equations
we already have, but change notation slightly:
Replace a by g = 9.8 m/s2
But in the equations it could have
a + or a – sign in front of it!
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 NEVER is negative!!!
The sign (+ or -) of the
gravitational acceleration is taken
into account in the equations we
now discuss!
The Sign of g in 1d Equations
The magnitude (size) of g = 9.8 m/s2
(A POSITIVE NUMBER!)
• But, acceleration is a vector (1 dimen), with 2
possible directions. Call these + and -.
• However, which way is + and which way is – is
ARBITRARY & UP TO US!
• May seem “natural” for “up” to be + y and
“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.
• 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. Acceleration 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!
Equations for Objects in Free Fall
• Written taking “up” as + y!
v = v0 - gt
(1)
y = y0 + v0 t – (½)gt2 (2)
2
2
v = (v0) - 2g (y - y0) (3)
v = (½)(v + v0)
(4)
g = 9.8 m/s2
Usually y0 = 0. Sometimes v0 = 0
Equations for Objects in Free Fall
• Written taking “down” as + y!
v = v0 + gt
(1)
y = y0 + v0 t + (½)gt2 (2)
2
2
v = (v0) + 2g (y - y0) (3)
v = (½)(v + v0)
(4)
g = 9.8 m/s2
Usually y0 = 0. Sometimes v0 = 0
Example: Falling from a Tower
A ball is dropped (v0 = 0) from a tower 70.0 m high. How
far will it have fallen after time t1 = 1 s, t2 = 2 s, t3 = 3 s?
Note: y is positive
DOWNWARD!
v = gt, y = (½) gt2
a = g = 9.8 m/s2
v1 = (9.8)(1)
= 9.8 m/s
v2 = (9.8)(2)
= 19.6 m/s
v3 = (9.8)(3)
= 29.4 m/s
Example: Thrown Down From a Tower
A ball is thrown down with an initial downward velocity of
v0 = 3 m/s, instead of being dropped. What are it’s position &
speed after t1 = 1 s & t2 = 2 s? Compare with the dropped ball.
y is positive
DOWNWARD!
v = v0 + gt
y = v0t + (½)gt2
a = g = 9.8 m/s2
v1 = 3 + (9.8)(1) = 12.8 m/s
y1 = (3)(1) + (½)(9.8)(1)2
= 3.0 + 4.9 = 7.9 m
v2 = 3 + (9.8)(2) = 20.6 m/s
y2 = (3)(2) + (½)(9.8)(2)2
= 6.0 + 19.6 = 25.6 m
v2 = 3 + (9.8)(2) = 20.6 m/s
y2 = (3)(2) + (½)(9.8)(2)2
= 6.0 + 19.6 = 25.6 m
Example:
A person throws a ball upward into the
air with an initial velocity v0 = 15.0 m/s.
v = 0 here,
but a = - g!
Calculate:
a. The time to reach the maximum height.
b. The maximum height.
c. The time to come back to the hand.
d. The velocity when it returns to the hand.
Time to top
= ½ round
trip time!
vA = 15 m/s
 v C = - v0
= -15 m/s
Note: y is positive UPWARD!
v = v0 – gt, y = v0t - (½)gt2
v2 = (v0)2 - 2g(y - y0)
Work this in general on
the white board!
Example: A ball is thrown upward at an initial speed
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 upward 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.