Download d = 0.5 gt 2

Motion of a
Freely
Falling Body
Objectives
► Explain
Galileo’s theory of motion and
Newton’s laws of motion.
► Apply all the equation for motion of an
object in free fall.
► To solve problems on free falling bodies
and the acceleration of gravity.
Galileo
►The
remarkable
observation that all
free falling objects
fall at the same rate
was first proposed
by Galileo, nearly
400 years ago.
►Galileo
conducted experiments
using a ball on an inclined plane to
determine the relationship between
the time and distance traveled.
Galileo
►He
found that the
distance depended
on the square of the
time and that the
velocity increased as
the ball moved
down the incline.
►The
relationship was the same
regardless of the mass of the ball
used in the experiment.
► The
Galileo
story that Galileo
demonstrated his
findings by dropping two
cannon balls off the
Leaning Tower of Pisa is
just a legend.
► However, if the
experiment had been
attempted, he would
have observed that one
ball hit before the other!
Introduction to Free Fall
A free-falling object is an object which is
falling under the sole influence of gravity.
► That is to say that any object which is
moving and being acted upon only be the
force of gravity is said to be "in a state of
free fall."
► This definition of free fall leads to two
important characteristics about a freefalling object:
 Free-falling objects do not encounter air
resistance.
 All free-falling objects (on Earth)
►
Freely Falling Body
►
►
►
The acceleration of freely falling body
is so important that physicist called it
acceleration due to gravity.
Denoted by letter g which is
equivalent to 32ft/sec2 or 9.8m/sec2.
Meaning in the 1st second, a falling
body accelerates from a stationary
position to a velocity of 9.8m/sec2,
after 2 seconds, the velocity is
doubled to 19.6m/sec2 after 3
seconds it triples to 29.4m/sec2
Freely Falling Body
►
►
►
Since accelerating objects are
constantly changing their velocity,
you can say that the distance
traveled divided by the time taken to
travel that distance is not a constant
value.
A falling object for instance usually
accelerates as it falls.
The fact that the distance which the
object travels every interval of time
is increasing is a sure sign that the
ball is speeding up as it falls
downward
A simple rule to bear in mind
► is
that all objects
(regardless of their mass)
experience the same
acceleration when in a state
of free fall.
► When the only force is
gravity, the acceleration is
the same value for all
objects.
► On Earth, this acceleration
value is 9.8 m/s/s that it is
given a special name - the
Newton
►Newton's
first law the law of
interaction states
that for every
action there's an
equal and opposite
reaction.
Newton
► Newton's
second law the law of
acceleration states
that the acceleration
of an object is directly
related to the net
force and inversely
related to its mass.
Newton’s Law of Acceleration
► Fnet
=m*a
►A=F/m
Acceleration depends upon two
factors: force and mass.
► The
10-kg elephant obviously
has more mass (or inertia).
This increased mass has an
inverse effect upon the
elephant's acceleration.
► And thus, the direct effect of
greater force on the 10-kg
elephant is offset by the
inverse effect of the greater
mass of the 10-kg elephant;
and so each object accelerates
at the same rate approximately 10 m/s/s.
One Newton is defined as the amount of
force required to give a 1-kg mass an
acceleration of 1 m/s/s.
Complete the table
Net Force
(N)
Mass
(kg)
Acceleration
(m/s/s)
1.
10
2
5 m/s/s
2.
20
2
10 m/s/s
3.
20
4
5 m/s/s
4.
10
2
5
5.
10
1
10
Free Fall and the Acceleration of
Gravity
► Free-falling
objects are in a state
of acceleration.
► The velocity of a free-falling object
is changing by approximately 10
m/s every second.
► If dropped from a position of rest,
the object will be traveling
approximately 10 m/s at the end
of the first second, approximately
20 m/s at the end of the second
second, approximately 30 m/s at
the end of the third second, etc.
Free Fall and the Acceleration of
Gravity
► The
velocity of a freefalling object which has
vf
=
g
t
been dropped from a
position of rest is
► where g is the
dependent upon the
acceleration of
time for which it has
gravity. An
fallen.
approximate
value
► The formula for
for g on Earth is
determining the
10 m/s/s; more
velocity of a falling
exactly, its value
object after a time
is 9.8 m/s/s.
of t seconds is
Example calculations for velocity
►Calculate
the
velocity of a freefalling object
after six, and
eight seconds
►
vf = g t
► Solution
► At
t=6s
► vf = (10 m/s2) (6 s)
►
= 60 m/s
► At
t=8s
► vf = (10 m/s2) (8 s)
►
= 80 m/s
Velocity of Freely Falling Body
►
►
►
►
►
If you were to observe the
motion of a free-falling
object you would notice that
the object averages a
velocity of
5 m/s in the first second,
15 m/s in the second
second,
25 m/s in the third second,
35 m/s in the fourth second,
etc.
Distance
► The
distance which a free-falling
object has fallen from a position
of rest is also dependent upon
the time of fall.
► This distance can be computed
by use of a formula; the
distance fallen after a time of t
seconds is given by the formula.
d = 0.5 g t2
► where g is the acceleration of
gravity
Example calculations for distance
►
► Calculate
the
distance fallen by a
free-falling object
after one, two and
five seconds
►
►
►
►
►
►
►d
= 0.5gt2
►
►
►
Example Calculations:
At t = 1 s
d = (0.5) (10 m/s2) (1 s)2
=5m
At t = 2 s
d = (0.5) (10 m/s2) (2 s)2
= 20 m
At t = 5 s
d = (0.5) (10 m/s2) (5 s)2
= 125 m
Distance
Given these average velocity
values during each consecutive
1-second time interval, the
object falls:
► – 5 meters in the first second,
– 15 meters in the second
second (for a total distance of
20 meters),
– 25 meters in the third second
(for a total distance of 45
meters),
– 35 meters in the fourth second
(for a total distance of 80
meters).
►
d=0.5(10g/s2 )(1s)2
v=0.5(10m/s2 )1s
Time
Interval
Average Velocity
During Time
Interval
v=0.5gt
d= 20 -5
Total Distance
Distance
Traveled from 0 s
Traveled During
to End of Time
Time Interval
Interval
d=dt -di
d=0.5gt2
0-1s
5 m/s
5m
5m
1-2s
15 m/s
15 m
20 m
2-3s
25 m/s
25 m
45 m
3-4s
35 m/s
35 m
80 m
The table illustrates that a free-falling object
which is accelerating at a constant rate will
cover different distances in each consecutive
second.
Further analysis of the first and last columns of
the table above reveal that there is a square
relationship between the total distance traveled
and the time of travel for an object starting
from rest and moving with a constant
acceleration.
For objects with a constant acceleration, the
distance of travel is directly proportional to the
square of the time of travel.
Equation for motion of an
object in free fall:
d
v
t
v  vo
g
t
v  vo  gt
v f  gt
d  v0t  0.5 gt
v  v  2 gd
2
2
0
 v0  v 
d 
t

 2 
d  0.5 gt
2
2
Sample Problem
►A
coin was dropped from the top
of the LTA building with a height of
727 ft. If there is no air resistance,
how fast (ft/s) will the coin be
moving when it hits the ground?
►215.68 ft/s
Solution A
d  0.5 gt
t  45.44
2
d  0.5(32 ft / s )t
2
t  6.74
2
727 ft  0.5(32 ft / s )t
2
(727 ft )
2
t 
2
16 ft / s
2
v  at
v  (32 ft / s )(6.74s )
2
v  215.68 ft / s
Problem 1
►A
marble is dropped from a bridge
and strikes the water in 5 seconds.
Calculate the speed with which it
strikes and the height of the
bridge.
► (Vf
= 49 m/s, d = 122.5 m)
Solution 1
v f  gt
v f  (9.8m / s )(5s )
2
v f  49m / s
d  0.5 gt
2
d  0.5(9.8m / s )(5s )
2
d  122.5m
2
P2
►A
feather is dropped on the moon
from a height of 1.40 meters. The
acceleration of gravity on the
moon is 1.67 m/s2. Determine the
time for the feather to fall to the
surface of the moon.
► T = 1.29 sec
S2
Given :
v i  0m/s
d  1.40m
a  1.67m/s
t?
2
P3
►The
observation deck of the World
Trade Center is 420 m above the
street. Determine the time required
for a penny to free fall from the
deck to the street below.
►T = 9.26 sec
S3
Given
vi  0m / s
d  0.5gt
2
 420m  0.5(9.8m / s )t
2
 420m  (4.9m / s )t
d  420m
 420m
2

t
2
g  9.8m / s  4.9m / s 2
2
2
t ?
85.7 s  t
2
t  85.7 s  9.26s
2
2
2
P4
►With
what speed in miles/hr
(1 m/s = 2.23 mi/hr) must an
object be thrown to reach a height
of 91.5 m (equivalent to one
football field)? Assume negligible
air resistance.
►V = 94.4 mi/hr
P4
v  v  2 gd
2
f
2
i
0m / s 
2
Given :
 v  2(9.8m / s )(91.5m)
2
i
2
g  9.8m / s 0  v  1793m / s
v f  om / s vi2  1793m 2 s 2
d  91.5m
vi  1793m / s  42.3m / s
vi  ?
2
t ?
2
i
2
2
(42.3m / s)( 2.23mi / hr )
vi 
1m / s
vi  94.4mi / hr
P5
►A
10kg block being held at rest above
the ground is released. The block
begins to fall under only the effect of
gravity. At the instant that the block
is 2.0 meters above the ground, the
speed of the block is 2.5m/sec. The
block was initially released at a height
of how many meters.
►D = 2.3 m
v  v  2 g (d  d 0 )
2
S5
2
0
v0  0
d 0  initialhei ght v  2 g (d  d 0 )
2
v0  0
d  2m
v  2.5m / s
m  10kg
g  9.8m / s
2
v
 d  d0
2g
2
0.5v
d0 
d
g
2
0.5(2.5m / s )
d0 
 2m
9.8m / s / s
d  2.3m
Assignment
1. Miguel drops a pile of roof
shingles from the top of a roof
located 8.52 meters above the
ground. Determine the time
required for the shingles to reach
the ground.
► 2. Brandy throws his mother's
crystal vase vertically upwards with
an initial velocity of 26.2 m/s.
Determine the height to which the
vase will rise above its initial height.
►
Assignment
3. A kangaroo is capable of jumping to
a height of 2.62 m. Determine the
take-off speed of the kangaroo.
► 4. A stone is dropped into a deep well
and is heard to hit the water 3.41 s
after being dropped. Determine the
depth of the well.
► 5. Ronald McDonald is riding an Air
Balloon on his way to Subic. If Ronald
free-falls for 2.6 second, what will be
his final velocity and how far will he
►
► Prepare
► END
for an EXAM next meeting….
OF PRESENTATION
►
A bowling ball falls
freely (near the
surface of the
Earth) from rest.
How far does it fall
in 4 seconds, and
how fast will it be
going at that time?
The ball's average speed for the first 4 seconds is the
average of 0 m/s and 40 m/s, its starting and ending
speeds, and distance = average speed times time.
► So the object will have fallen 80 meters, and its
speed will be 40 m/s.
►
►
The position of a free-falling body (neglect air
resistance) under the influence of gravity can be
represented by the function
1 2
s (t )  gt  v0t  s0
2
where g is the acceleration due to gravity (on
earth ) and s0 and v0 are the initial height and
velocity of the object (when ).
► Example
(Falling Object Problem) A ball is
thrown vertically upward from the ground with
an initial velocity of 160 ft/s.
(a) When will it hit the ground?
Solution. Since with and we need
with
and
we need
This is precisely when or and thus the ball will
hit the ground in 10 seconds.
This illustrates that a free-falling object which is
accelerating at a constant rate will cover
different distances in each consecutiveTotal
second.
Distance
Average Velocity
Distance Traveled
Traveled from 0 s
Further
and
Time
Interval analysis
During of
Timethe first
During
Timelast columns of
to End of Time
Interval
Interval
the table above reveal that there is aInterval
square
relationship between the total distance traveled
the time
0 - 1and
s
5 m/sof travel for
5 m an object 5starting
m
from rest
and moving
with a constant
1-2s
15 m/s
15 m
20 m
acceleration.
2-3
m/s
25 m
45 m
For objects25 with
a constant
acceleration,
3-4s
35 m/s
35 m
80 m
the distance
of travel
is directly
proportional to the square of the time of
travel.
►
►
►
As such, if an object travels for twice the time, it will
cover four times (22) the distance; the total distance
traveled after two seconds is four times the total distance
traveled after one second.
If an object travels for three times the time, then it will
cover nine times (32) the distance; the distance traveled
after three seconds is nine times the distance traveled
after one second.
Finally, if an object travels for four times the time, then it
will cover sixteen times (42) the distance; the distance
traveled after four seconds is sixteen times the distance
traveled after one second.
►
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