Free Fall
Are you an Aristotle?
or A Galilei'n?
Aristotle believed that "heavier" objects fell faster Galileo Galilei believed that all objects fell simultaneously
Will they fall at the same
speed?
David Scott's lunar
experiment tested this
theory
experiment by David Scott on lunar surface Apollo 15
Galileo (1564- 1642) disagreed
with Aristotle (384-322 BC)
Galileo studied falling objects using
an inclined plane... in a way he
slowed object down much like we
did in the lab. He changed the
weight of the "cart" and his data
showed that all objects regardless
of mass fell the same way...
They had the same acceleration
Hi Everyone!
Take the notes and try the problems
We can go over questions after the break
Have a great March Break!!!!!!!
See you in March
Get some rest and have some fun!!!!!
Mrs. Martin
Experiment in the Foyer
Last class I asked you to find the acceleration
of 6 balls.
You were given their average time they were
dropped 10 times and timed each time.
We concluded that the accelerations were very
close, some reasons for variation might be
human error and air resistance. This can be
affected by shape for example.
Definition of a free falling object:
A freely falling object is any object moving freely under
the influence of gravity alone, regardless of its initial
motion.
Objects thrown upward or downward and those
released from rest are all falling freely once they are
released.
Any freely falling object experiences an acceleration
directed downward, regardless of its initial motion.
motion of free falling objects
Direction is vertical, up(+) and down(-) +
xi= point of reference
which we usually make 0 m.
xi
-
Acceleration due to gravity is directed
downward therefore g is -
"g" = Acceleration of free fall
g = -9.8 m/s2
(at the earth's surface)
Initial velocity will be positive when we throw something up
Initial velocity will be negative when we throw something down
Initial velocity is zero when we just let go....
So, because it is rectilinear (this time
vertical instead of horizontal)
We use the same formulas for UARM
(the 4 formulas on the board) and a is
now g (-9.8m/s2)
Example:
A stone thrown from the top of a building is given an initial
velocity of 20.0 m/s straight upward. The building is 50.0 m
high, and the stone just misses the edge of the roof on its way
down, as shown on the next slide. Using ti= 0s as the time the
stone leaves the thrower's hand at position A, determine
write down what you know
x1 = 0 (at the top of the building, where the ball is thrown)
vi = 20.0 m/s
a = g = -9.8m/s2
ti = 0s
A) the time at which the stone reaches its maximum height A TO B
t = ? vf = 0m/s
B) the maximum height
Δx = ? when v
C) i-the time at which the stone returns to the
height from which it was thrown
ii- the velocity of the stone at this instant (
D) The velocity and position at t = 5.00s
Draw it out.... (next slide)
B
Draw the situation
a) A to B: how long?
vi = +20 m/s
a = -9.8 m/s2
ti = 0
A
C
D
E
Forumula a = -9.8m/s2
a) vf= vi + at
0 = 20m/s + -9.8m/s2(t)
-20m/s ÷ -­9.8m/s2 = t
t = 2.04s
b)
xf = xi + vit + 1/2(g)(t2)
xf = 0 + (20m/s)(2.04s) + 1/2(-9.8m/s2)(2.04)2
xf = 20.4m
B
c) time to get to C (the stone returns to the height it was thrown
A
C
vi = 20 m/s
a = -9.8 m/s2
ti = 0
tf = ?
t = ?
V2 = ?
D
It returns in the same time so 2 x 2.04s
= 4.08s
E
formula: Or using the formula:
2
Δy = v1t + 1/2 at
0 = 20.0t ­ 4.90t2
Factor out t
0 = t(20 ­ 4.9t)
t = 0 or 0 = 20 ­ 4.9t
4.9 t = 20
t = 4.08s
so t either = 0 (the starting time)
or 4.08s when it returns to x= 0
B
Draw the situation
vi = 20 m/s
a = -9.8 m/s2
ti = 0
tf = 2.04s
A
C
D
velocity at t = 4.08 s
E
formula: v2 = v1 + at
v2 = 20.0m/s + ­9.8m/s2(4.08s)
= ­20 m/s
B
Draw the situation
A
vi = 20 m/s
a = -9.8 m/s2
ti = 0
tf = ?
xf = -50m
formula: C
D
E
xf = xi + v1t + 1/2 at2
­50 = 0 + v1 t + ­50m = 0m + 20m/s (t) + 1/2 (­9.8m/s2)(t2)
v2 = 0 + gt = 20m/s(5s) + ­9.8m/s2(5.00s)2
v2 = ­29 m/s
B
Draw the situation
A
vi = 20 m/s
a = -9.8 m/s2
ti = 0
tf = 5s
C
D
E
formula: v2 = v1 + gt = 20m/s + ­9.8m/s2(5.00s)
v2 = ­29 m/s
Practice:
Handout given February 18th
New Text p238
old text p 225 (answers in back)
Then chapter review:
p 242 questions 1, 3, 4, 5, 7, 8, 9
Enriched(optional): #10: it's a quadratic: use the quadratic formula
Answers will be posted on the site.
Attachments
experiment by David Scott on lunar surface Apollo 15