Free Fall
 Student determine the effect of gravity on objects
without support.
 Students will calculate these effects of gravity over
time.
1
Four Basic Equations of Physics
Vf = Vi + at
d = Vi t + ½ at2
2ad = Vf2 – Vi2
 d =Vf + Vi
2
t
2
Free Fall
Free fall – motion under
the influence of the
gravitational force only
(neglects air resistance)
3
Elapsed time
time that has passed
from the beginning
of a fall
4
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
5
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?
6
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=1s
7
Therefore…
G: a = 9.8 m/s2
Vi = 0
t=1s
U: Vf = ?
E: Vf = Vi + at
S: Vf = 0 + (9.8 m/s2) (1 s)
S: Vf = 9.8 m/s
8
Now find out how fast it will be
falling after 2 s…
 G: a = 9.8 m/s2
Vi = 0
t=2s
U: Vf = ?
E: Vf = Vi + at
S: Vf = 0 + (9.8 m/s2) (2 s)
S: Vf = 19.6 m/s
9
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
10
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
11
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?
12
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
13
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
14
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
2
gt
15
Acceleration due to Gravity
 If the object is moving down, a = 9.8 m/s2
a = 9.8 m/s2
(speeding up)
 If the object is moving up, a = - 9.8 m/s2
a = - 9.8 m/s2
16
(slowing down)
Throwing an object into the air
 When an object is thrown into the air, the velocity at its highest
point is ZERO!!!
End
Begin
Vf = 0 m/s
Vi = 0 m/s
Begin
End
Vi = ?
Vf = ?
17
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?
18
How high does the ball rise?
 HINT: draw a picture and label it
End
Vf = 0 m/s
d= ?
Begin
Vi = 4 m/s
19
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 = Vf2 – Vi2, (solve for d) d = Vf2 – Vi2
2a
S: d = 02 – (4 m/s)2
2(-9.8 m/s2 )
S: d = 0.816 m
20
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)
S: t = 0.408 s
21
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
22
What is its velocity just before it
is caught?
 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 = Vf2 – Vi2, (solve for V ) Vf2 = 2ad + Vi2
S: Vf2 = 2(9.8 m/s2 )(0.816 m) + 02
S: Vf = 3.99 or 4 m/s
f
23
You’re done!
Now try some
problems on your
own.
24
Keep your projects at your seat for now.
Place your completed Lab in the TOP tray.
Someone pass out all papers in the 4A
drawer.
Complete the warm-ups for today.