1.
Free Fall
Free fall – motion under
the influence of the
gravitational force only
(neglects air resistance)
1
2.
Elapsed time
time that has passed
from the beginning of
a fall
2
3.
Gravity and Free fall
Acceleration due to gravity
is 9.8 m/s2
, downward
Every second that an
object falls, the velocity
increases by 9.8 m/s
3
4.
Sample 1
If an object is dropped from rest at the
top of a cliff, how fast will it be going
after 1 second?
2 seconds?
10 seconds?
4
5.
First, Identify the GIVENS?
When an object is falling, assume that
a = 9.8 m/s2
“from rest” tells us that Vi = 0
“how fast will it be going?” is asking us
to find Vf = ? (this is the unknown)
the first problem gives us a time of
t = 1 s
Work this on YOUR paper using
GUESS… 5
6.
Therefore…
G: a = 9.8 m/s2
Vi = 0
t = 1 s
U: Vf = ?
E: Vf = Vi + at
S: Vf = 0 + (9.8 m/s2)
(1 s)
S: Vf = 9.8 m/s
6
7.
Now find out how fast it will be
falling after 2 s…
G: a = 9.8 m/s2
Vi = 0
t = 2 s
U: Vf = ?
E: Vf = Vi + at
S: Vf = 0 + (9.8 m/s2)
(2 s)
S: Vf = 19.6 m/s
7
8.
Now find out how fast it will be
falling after 10 s…
G: a = 9.8 m/s2
Vi = 0
t = 10 s
U: Vf = ?
E: Vf = Vi + at
S: Vf = 0 + (9.8 m/s2)
(10 s)
S: Vf = 98 m/s
8
9.
Calculating How Fast Free
Fall Is
If we use Vf = Vi + at, and Vi = 0 then we actually
have an equation that reads… Vf = at, where a = 9.8
m/s2
. When a = 9.8 m/s2
, we call that constant g.
Therefore, we can use the following new, or derived
equation…
Instantaneous speed = acceleration X elapsed time
v = gt
9
10.
Sample Problem #2
The Demon Drop ride at Cedar
Point Amusement Park falls
freely for 1.5 s after starting
from rest.
What is its velocity at the end of
1.5 s?
How far does it fall?
10
11.
What is its velocity at the end
of 1.5 s?
G: a = 9.8 m/s2
Vi = 0 (starting from rest)
t = 1.5 s
U: Vf = ?
E: Vf = Vi + at
S: Vf = 0 + (9.8 m/s2)
(1.5 s)
S: Vf = 14.7 m/s
11
12.
How far does it fall?
G: a = 9.8 m/s2
Vi = 0
t = 1.5 s
Vf = 14.7 m/s (from previous part)
U: d = ? (how far)
E: d = Vit + ½ at2
(use any of the 3 motion formulas with d in
them)
S: d = (0)(1.5s) + ½ (9.8 m/s2
)(1.5s)2
S: d = 11.03 m
12
13.
Calculating How Far Free
Fall Is
If we use d = Vit + ½ at2
, and Vi = 0 then we
actually have an equation that reads…
d = ½ at2
, where a = 9.8 m/s2
. When a = 9.8
m/s2
, we call that constant g. Therefore, we
can use the following new, or derived
equation…
d = ½ gt2
13
14.
Acceleration due to Gravity
If the object is moving down, a = 9.8 m/s2
If the object is moving up, a = - 9.8 m/s2
a = 9.8 m/s2
(speeding up)
a = - 9.8 m/s2
(slowing down)
14
15.
Throwing an object into the air
When an object is thrown into the air, the
velocity at its highest point is ZERO!!!
Begin
Vi = ?
Begin
Vi = 0 m/s
End
Vf = 0 m/s
End
Vf = ?
15
16.
Sample Problem #3
A ball is thrown vertically into the air with an
initial velocity of 4 m/s.
How high does the ball rise?
How long does it take to reach its highest point?
If the ball is caught in the same spot from which
it was thrown, what is the total amount of time
that it was in the air?
What is its velocity just before it is caught?
16
17.
How high does the ball rise?
HINT: draw a picture and label it
Begin
Vi = 4 m/s
End
Vf = 0 m/s
d= ?
17
18.
How high does the ball rise?
G: a = -9.8 m/s2
(notice the negative sign, ball moving upward)
Vi = 4 m/s
Vf = 0 m/s (at the top, before it starts to fall, it stops)
U: d = ?
E: 2ad = Vf
2
– Vi
2
, (solve for d) d = Vf
2
– Vi
2
2a
S: d = 02
– (4 m/s)2
2(-9.8 m/s2
)
S: d = 0.816 m
18
19.
How long does it take to reach its
highest point?
G: a = -9.8 m/s2
U: t = ?
Vi = 4 m/s
Vf = 0 m/s
d = 0.816 m
E: Vf = Vi + at, (solve for t) t = Vf – Vi
a
S: t = 0 m/s – 4 m/s
(-9.8 m/s2
)
19
20.
If the ball is caught in the same spot
from which it was thrown, what is
the total amount of time that it was
in the air?
G: a = 9.8 m/s2
(ball going down, positive) U: t = ?
Vi = 4 m/s
Vf = 0 m/s
d = 0.816 m
E: d = Vit + ½ at2
; derived to d = ½ gt2
, solve for t…
t2
= (2d)/g
S: t2
= (2)(0.816 m)/(9.8 m/s2
) *don’t forget to take the square root
S: t = 0.408s
Total time = time Going up + time going down
Total time = 0.408s +0.408s = 0.816s 20
21.
G: a = 9.8 m/s2
(ball moving down before it is caught)
Vi = 0 m/s
d = 0.816 m
t = 0.408 s
U: Vf = ?
E: 2ad = Vf
2
– Vi
2
, (solve for Vf) Vf
2
= 2ad + Vi
2
S: Vf
2
= 2(9.8 m/s2
)(0.816 m) + 02
S: Vf = 3.99 or 4 m/s
What is its velocity just before
it is caught?
21
22.
You’re done!
Now try some
problems on your
own.
22
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