Acceleration Due to Gravity
•EQ:
How is the motion of an
object moving vertically
different from one moving
horizontally?
Acceleration Due to Gravity
• In physics, a uniformly accelerated
motion in a vertical line is called free
fall.
• Every object on the earth experiences a
common force: the force due to gravity.
• This force is always directed toward the
center of the earth (downward).
• Free fall can be analyzed using the same
•
kinematic equations we used to analyze
horizontal motion. But there are a few factors
we must take into consideration.
http://www.youtube.com/watch?v=_d8ROhH3_vs
Gravitational Acceleration
• In a vacuum, all objects fall with same
acceleration.
• Equations for constant acceleration
apply as usual.
• Near the Earth’s surface:
a = g = - 9.80 m/s2
Directed downward (always negative).
The Acceleration of Gravity (g)
• If two rocks with different mass are
dropped (vi = 0) from the same height at
the same time, would one land before the
other? Or would they land at the same
time?
• Galileo demonstrated that g is the same
for all objects, regardless of their mass!
• This was confirmed by the Apollo
astronauts on the Moon, where there is no
air resistance
• http://www.youtube.com/watch?v=7eTw
35ZD1Ig
Falling objects accelerate at a
constant rate (Galileo):
• All objects (regardless of their mass) experience
the same acceleration when free falling. No air
resistance present.
• When the only force is gravity, the acceleration
is the same for all objects. On Earth, this
acceleration (g) is 9.8 m/s2.
• When an object is dropped or simple falls, it will
have an vi = 0 m/s since no initial force was
applied to make it move.
Gravitational Acceleration
• When an object is thrown straight up, it will have a velocity of 0 m/s2
at its peak height just before it begins its free fall.
–This means that for a split second, an object will stop in mid air
while it transitions from upward motion to downward motion.
• Because gravity is constant, the rate at which an object decelerates
as it moves upward, is the same rate it accelerates at as it falls
downward.
–In other words, an object falling will have the same speed at any
point that it did on the way up.
6
Various Types of Free Fall
0 m/s
9.8 m/s
19.6 m/s
29.4 m/s
39.2 m/s
49.0 m/s
7
Graphing Free Fall Motion
Talk out your misconceptions about free
falling objects with a shoulder partner then
at the bottom of your notes explain:
-Why any 2 objects will hit the ground first regardless
of their mass (in the absence of air resistance)
-Why the initial velocity of free falling objects is zero
-What direction does gravity act on (side to side or up
and down)
-What does it mean to say that the acceleration of
gravity is 9.80 m/s2 (for ex: what is happening to the
object every second as it falls)
Sign Convention: A
Ball Thrown Vertically
Upward
x= +
v= 0
x= +
v= +
UP = +
Release Point
x= +
v= x= 0
v= -
x= v= -
Displacement is positive (+)
or negative (-) based on
LOCATION.
Velocity is positive (+) or
negative (-) based on
direction of motion.
Acceleration is gravity
which is always downward
(negative).
Horizontal Motion
v
d
t
Vertical Motion
v
d
t
vf = vi + a(Δt)
vf = vi + g(Δt)
Δx = ½ (vi + vf)Δt
Δy = ½ (vi + vf)Δt
Δx = vi(Δt) + ½ a(Δt)2
Δy = vi(Δt) + ½ g(Δt)2
vf2 = vi2 + 2aΔx
vf2 = vi2 + 2gΔy
Drop/fallVi=0 m/s
Height, tall Δy
g= acceleration due
to gravity = -9.81
m/s2
Same Problem Solving Strategy
Except a = g:
 Draw and label sketch of problem.
 Indicate + direction.
 List givens and state what is to be found.
 Select equation containing one and not
the other of the unknown quantities, and
solve for the unknown.
 Don’t Forget Units!!!!!
Example 7: A ball is thrown vertically upward with an
initial velocity of 30 m/s. What are its position and
velocity after 2 s, 4 s, and 7 s?
Step 1. Draw and label a
sketch.
Step 2. Indicate + direction.
Step 3. Identify given
variables.
a = -9.8 ft/s2
t = 2, 4, 7 s
vi = + 30 m/s y= ? v = ?
+
a=g
vi = +30 m/s
Finding Displacement:
Step 4. Select equation
that contains x and not v.
y  vi t 
1
2
at
2
+
a=g
x = (30)t + ½(-9.8)t2
Substitution of t = 2, 4, and 7 s
will give the following values:
vo = 30 m/s
y = 40.4 m; y= 41.6 m; y= -30.1 m
Finding Velocity:
Step 5. Find v from equation
that contains v and not x:
v f  vi  at
+
a=g
v f  (30)  (9.8)t
Substitute t = 2, 4, and 7 s:
vo = 30 m/s
v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s
Example 7: (Cont.) Now find the
maximum height attained:
Now find an equation that
uses y but not t. Keep in
mind the velocity at its
highest point is zero
+
a=g
v f  vi  2 gy
2
2
0  (30)  2(9.8) y
2
2
2
30
y
 45.9m
2(9.8)
vo = +30 m/s
The tallest Sequoia sempervirens
tree in California’s Redwood
National Park is 111 m tall.
Suppose an object is thrown
downward from the top of that
tree with a certain initial velocity.
If the object reaches the ground
in 3.80 s, what is the object’s
initial velocity?
Horizontal Motion
v
d
t
Vertical Motion
v
d
t
vf = vi + a(Δt)
vf = vi + g(Δt)
Δx = ½ (vi + vf)Δt
Δy = ½ (vi + vf)Δt
Δx = vi(Δt) + ½ a(Δt)2
Δy = vi(Δt) + ½ g(Δt)2
vf2 = vi2 + 2aΔx
vf2 = vi2 + 2gΔy
Drop/fallVi=0 m/s
Height, tall Δy
g= acceleration due
to gravity = -9.81
m/s2
G:
E:
S:
U:
vi
S: