Gravity by gjjur4356

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									Acceleration: Gravity


3U Physics
                                             g

The acceleration due to the Earth‘s gravity is
9.8 m/s2 [down] (or –9.8 m/s2).
                      32  1
The magnitude of this acceleration is denoted
  by the letter g.
             Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
               Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
Givens :
a  9.8 sm2


d   25 m
               Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
Givens :
a  9.8 sm2


d   25 m
v1  0
Unknown :
t  ?
                  Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
Givens :                d  v1t  1 at 2 becomes
                                    2


a   9 .8   m          d  1 at 2 for v1  0
                             2
             s2

d   25 m
v1  0
Unknown :
t  ?
                  Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
Givens :                  d  v1t  1 at 2 becomes
                                      2


a   9 .8   m            d  1 at 2 for v1  0
                               2
             s2

d   25 m                                  2d
                      d  at
                           1      2
                                       t 
v1  0                     2
                                              a
Unknown :                  2 25 m 
                      t 
t  ?                       9.8 sm2
                  Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
Givens :                  d  v1t  1 at 2 becomes
                                      2


a   9 .8   m            d  1 at 2 for v1  0
                               2
             s2

d   25 m                                  2d
                      d  at
                           1      2
                                       t 
v1  0                     2
                                              a
Unknown :                  2 25 m 
                      t              2.3 s
t  ?                       9.8 s 2
                                  m
             Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
                 Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
t  ?
a   9.8   m
            s2

vi  0
               Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
t  ?
a   9.8 sm
           2


vi  0
 d   25 m
               Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
t  ?            d  vi  t  2 a t 2
                                1


a   9.8 sm
           2


vi  0
 d   25 m
               Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
t  ?            d  vi  t  a t
                               1
                               2
                                       2


a   9.8 sm
           2     for vi  0,  d  2 a t 2
                                   1


vi  0
 d   25 m
               Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
t  ?            d  vi  t  a t
                               1
                               2
                                       2


a   9.8 sm
           2
                 for vi  0,  d  a t
                                   1
                                   2
                                           2

vi  0                  2 d
 d   25 m  t        a
               Example: a falling object
An object is dropped from the top of a 25-m tall
 building. How long does it take to hit the ground?
t  ?            d  vi  t  a t
                              1
                              2
                                       2


a   9.8 sm
           2
                 for vi  0,  d  a t
                                   1
                                   2
                                           2

vi  0                  2 d       2(  25 m)
 d   25 m  t        a
                             
                                     9.8 s2
                                           m   2.3 s
              Example: up, then down
An object is tossed upwards at 5.0 m/s. How long
 does it take to reach its maximum height?
               Example: up, then down
An object is tossed upwards at 5.0 m/s. How long
 does it take to reach its maximum height?
Givens :
a  9.8 sm2


v1  5.0 m
         s
               Example: up, then down
An object is tossed upwards at 5.0 m/s. How long
 does it take to reach its maximum height?
Givens :
a  9.8 sm2


v1  5.0 m
         s

v2  0
Unknown:
t  ?
               Example: up, then down
An object is tossed upwards at 5.0 m/s. How long
 does it take to reach its maximum height?
Givens :          v2  v1
               a
a  9.8 sm2        t
v1  5.0 m
         s

v2  0
Unknown:
t  ?
               Example: up, then down
An object is tossed upwards at 5.0 m/s. How long
 does it take to reach its maximum height?
Givens :          v2  v1          v 2  v1
               a            t 
a  9.8 sm2        t                 a
v1  5.0 m
         s

v2  0
Unknown:
t  ?
               Example: up, then down
An object is tossed upwards at 5.0 m/s. How long
 does it take to reach its maximum height?
Givens :          v2  v1          v 2  v1
               a            t 
a  9.8 sm2        t                 a
v1  5.0 m
         s
                    0  5.0 m
v2  0         t             0.51s
Unknown:              9.8 sm
                            2


t  ?
               An inconstant constant

g is a ―constant‖ that varies according to your
  location on Earth:
                         32  1



Toronto                           g = 9.805 m/s2
North Pole (sea level)            g = 9.832 m/s2
Equator (sea level)               g = 9.780 m/s2
                                           g’s

Accelerations are often given in terms of g.
For example,


                   1g 
       49   m
            s2
                           5g
                   9.8 s2 
                        m
                                Blackout

A typical person can handle about 5 g
 before loss of consciousness,
 ―blackout,‖ occurs.
The record for the most g forces on a
 roller coaster belongs to Mindbender
 at Galaxyland Amusement Park in
 Edmonton, Alberta, at 5.2 g.
                                       Greyout
Through the combination of special g-suits
  and efforts to strain muscles —both of
  which act to force blood back into the
  brain— modern pilots can typically
  handle 9 g or more. They may sometimes
  experience a greyout between 6 and 9 g.
  This is not a total loss of consciousness
  but is characterised by temporary loss of
  colour vision, tunnel vision, or an
  inability to interpret verbal commands
                              Negative g’s

Resistance to "negative" or upward g‘s, which
 drive blood to the head, is much less. This
 limit is typically in the -2 to -3 g range. The
 vision goes red and is also referred to as a
 "redout". This is probably due to capillaries
 in the eyes bursting under the increased
 blood pressure.
                                 “g-Force”

The human body is considerably more able to
 survive acceleration that is perpendicular to
 the spine. In general, when acceleration
 pushes the body backwards (colloquially
 known as 'eyeballs in') a much higher
 tolerance (17g) is shown than when the
 body is pushed forwards (‗eyeballs out,‘
 12g) since blood vessels in the retina appear
 more sensitive to that direction.
        Strongest g-forces survived

Voluntarily: Colonel John Stapp in 1954
 sustained 46.2 g in a rocket sled, while
 conducting research on the effects of human
 deceleration
         Strongest g-forces survived

Involuntarily: Formula One racing car driver
  David Purley survived an estimated 178 g in
  1977 when he decelerated from 173 km/h to
  0 in a distance of 66 cm after his throttle got
  stuck wide open and he hit a wall
     Everyday g-forces

Coughing: 3.5 g
Sneezing: 2.9 g
                            Free fall

Objects in free-fall feel
0 g, or ―weightlessness.‖
                   Mass doesn’t matter

Note that all objects, regardless of mass,
 experience the same acceleration.
                                    Galileo

This theory (contrary to what Aristotle had
 taught) is attributed to Galileo.
                                      Drag

However, some objects are slowed by
 atmospheric drag more than others.
         An equation for drag

Fd  v ACd
     1
     2
         2


Fd  drag force
  density of the atmosphere
v  relative speed of the object
A  reference area
Cd  drag coefficient
                      Terminal velocity

At a given speed, the drag force will equal the
 gravitational force, and the object will stop
 accelerating, i.e. reach ―terminal velocity.‖
                    Terminal velocities

Typical terminal velocities:

Human                          53 m/s (190 km/h)
Human with parachute           5 m/s (18 km/h)
Dandelion seed                 0.5 m/s (1.8 km/h)
                         The fastest man

On August 16th, 1960 U.S. Air Force Captain
 Joe Kittinger broke the sound barrier (1240
 km/h) during a free-fall from the high
 altitude balloon Excelsior III, at an altitude
 of approximately 31 km.
                   Highest fall survived
                  (without a parachute)

Flight attendant Vesna Vulovič fell 10,000 m
  on January 26, 1972 when she was aboard a
  plane that was brought down by explosives
  over the Czech Republic.
She suffered a broken skull, three broken
  vertebrae (one crushed completely), and
  was in a coma for 27 days, but she
  survived!

								
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