Free Fall
recognize
equations of
motion
Equations of Motion
Displacement (Δx):
Velocity (v):
Acceleration (a):
change in position
displacement during time
change in velocity
𝑣! = 𝑣" + 𝑎𝑡
equations of motion
constant acceleration
1 #
∆𝑥 = 𝑣" 𝑡 + 𝑎𝑡
2
𝑣!#
=
𝑣"#
+ 2𝑎∆𝑥
1
∆𝑥 = 𝑣! + 𝑣" 𝑡
2
Δx
m/s
v
∆𝑥 = 𝑥! − 𝑥"
positive and negative
signs related to direction.
stop: 𝑣! = 0
rest: 𝑣" = 0
Δt
Δv
a
Δt
∆𝑣 = 𝑣! − 𝑣"
Equations of Motion
Volva comes villa at morning
𝑣!
=
𝑣" +𝑎𝑡
𝑣! = 𝑣" + 𝑎𝑡
Delta becomes vital at square half evening
1
#
∆𝑥
=
𝑣" 𝑡 + 𝑎𝑡
2
Volva double comes violet villa with 2 axes
𝑣! #
=
𝑣" # +
memorize
equations of
motion
2𝑎∆𝑥
Delta comes to half Volva villa tower
1
∆𝑥
=
𝑣! + 𝑣"
𝑡
2
1 #
∆𝑥 = 𝑣" 𝑡 + 𝑎𝑡
2
𝑣!# = 𝑣"# + 2𝑎∆𝑥
1
∆𝑥 = 𝑣! + 𝑣" 𝑡
2
use equations
of motion to
solve problems
Equations of Motion
A car with an initial speed of 6 m/s accelerates at uniform rate of
2 m/s2 for 4 seconds. Find:
final speed of the car?
displacement during time interval?
𝑣" = 6
𝑎=2
𝑣! = 𝑣" + 𝑎𝑡
=6+2 4
= 14 m/s
𝑡=4
𝑣! =? ?
∆𝑥 =? ?
1 #
∆𝑥 = 𝑣" 𝑡 + 𝑎𝑡
2
$
= 6 4 + # 2(4)#
= 24 + 16 = 40 m
use equations
of motion to
solve problems
Equations of Motion
A train travelling initially at 7 m/s accelerates uniformly at rate of
1 m/s2 for a distance of 25.5 m. what is the velocity at the end of
this distance?
𝑣" = 7
𝑎=1
𝑣!# = 𝑣"# + 2𝑎∆𝑥
= 7# + 2 1 25.5
= 49 + 51
= 100
∆𝑥 = 25.5
𝑣! =? ?
𝑣!# = 100
𝑣! = 100
= 10 m/s
Free Falling
When a ball is freely thrown from some height, it
moves downward till reaches the ground.
describe
free falling
air
resistance
When a ball is thrown upward with initial velocity, it
goes up to reach certain height then falls downward.
In both cases the forces acting on the ball are:
gravitational force and air resistance.
If we consider one dimension motion and ignore the
air resistance force then the only force acting on the
ball is Earth gravitational force, therefore the motion is
called: Free Falling.
gravitational
force
gravitational
force
equations of
motion for
free falling
Free Falling (Dropping)
If we drop an object or throw it downward with
initial velocity it goes downward directly under
influence of earth gravitation force (weight).
vi
motion
As object moving downward velocity is negative.
also object gains negative acceleration due to
Earth’s gravity.
the object speeding up (velocity increases) as it
moving downward, till it reaches maximum
velocity before it hits the ground.
vf
a
equations of
motion for
free falling
Free Falling (Throwing)
If we throw an object upward with initial velocity
it goes up for a certain height then stops for
instant then falls down.
vf
As object moving upward velocity is positive.
also the acceleration is negative.
the object slowing down (velocity decreases) as it
moving upward, till it stops (zero).
a
motion
Even when the object moving upward it still
affected by gravitation force (weight).
vi
equations of
motion for
free falling
Free Falling
The object has negative acceleration equal to acceleration of gravity
𝑎 → −𝑔
∆𝑥 → ∆𝑦
𝑔 = 10 m/s #
𝑣! = 𝑣" + 𝑎𝑡
𝑣! = 𝑣" − 𝑔𝑡
1 #
∆𝑥 = 𝑣" 𝑡 + 𝑎𝑡
2
1 #
∆𝑦 = 𝑣" 𝑡 − 𝑔𝑡
2
𝑣!# = 𝑣"# + 2𝑎∆𝑥
𝑣!# = 𝑣"# − 2𝑔∆𝑦
1
∆𝑥 = 𝑣! + 𝑣" 𝑡
2
1
∆𝑦 = 𝑣! + 𝑣" 𝑡
2
equations of
motion
for free falling
regardless
of object’s
mass!!
Red Bull Stratos Project*
On 14 October 2012,
Felix Baumgartner flew
39 km in a helium
balloon before free
falling in a pressure
suit and then
parachuting to Earth.
his total jump, lasted approximately ten minutes.
While the free fall lasted 4 minutes and 19 seconds.
Baumgartner broke the sound barrier on his descent.
use equations
of motion to
solve problems
Free Falling
A boy drops a stone off the edge of a 125 m high.
Find velocity when the stone hits the ground?
Find time required to hit the ground?
𝑣" = 0
∆𝑦 = −125
𝑣!# = 𝑣"# − 2𝑔∆𝑦
𝑣! =? ?
𝑡 =? ?
𝑣! = 𝑣" − 𝑔𝑡
= 0# − 20(−125)
−50 = 0 − 10𝑡
= 2500
50
𝑡=
=5s
10
𝑣! = 2500 = −50 m/s
𝑦! = 125
𝑦" = 0
why negative
final velocity
use equations
of motion to
solve problems
Free Falling
A model rocket is launched straight upward with an
initial speed of 50 m/s. ignore air resistance
How high does the rocket go
How long does it take to reach that height
𝑣" = 50
𝑣! = 0
𝑣!# = 𝑣"# − 2𝑔∆𝑦
∆𝑦 =? ?
𝑣" = 0
𝑡 =? ?
𝑣! = 𝑣" − 𝑔𝑡
0 = 50# − 20(∆𝑦)
0 = 50 − 10𝑡
2500
∆𝑦 =
= 125 m
20
50
𝑡=
=5s
10
𝑣! = 50
why zero
final velocity
explain free
falling graphs
Free Falling
From equations of motion we can draw position – time graph and
velocity – time graph.
∆𝑦(m)
𝑎(m/s # )
𝑣" (m/s)
t(s)
t(s)
t(s)
𝑣!
−𝑔
the line on negative
y – axis because the
motion is downward
object is dropped.
the line on negative
y – axis because the
motion is downward
here 𝑣! ≠ 0.
the line on negative
y – axis because the
motion is downward
and its constant.
describe
free falling
Free Falling
Lets study the motion of object thrown upward then falls downward
suppose ball thrown with initial velocity of 40 m/s
time final velocity acceleration
1
+ 30
‒ 10
2
+ 20
‒ 10
3
+ 10
‒ 10
4
0
‒ 10
5
‒ 10
‒ 10
6
‒ 20
‒ 10
7
‒ 30
‒ 10
10
20
𝑣! = 30
The figure drawn in 2D to simplify the explanation!
0
−10
−20
−30
describe
free falling
Free Falling
From the table we notice few things:
Free falling objects undergo acceleration
downward and upward motion.
Free falling objects undergo constant
downward acceleration.
0
10
20
−10
−20
The velocities magnitude is the same at
the same height.
At maximum height the objects stop for a while then fall down.
Velocity could be positive or negative depending on direction of
motion. But the acceleration is always negative.
equations of
motion for
free falling
Free Falling
In free falling the gravitational
force the only force affects the
object motion, and the object
has acceleration equal to
acceleration of gravity
𝑣" = 0
𝑎 = −10
𝑣+
𝑣−
𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦
𝑎 = −10
𝑣! = 𝑣" − 𝑔𝑡
1 #
∆𝑦 = 𝑣" 𝑡 − 𝑔𝑡
2
𝑣!# = 𝑣"# − 2𝑔∆𝑦
𝑎 = −10
𝑦! = 0
equations of motion
constant acceleration
for free falling
𝑦" = 0
𝑎 → −𝑔
∆𝑥 → ∆𝑦
𝑔 = 10 m/s #
use equations
of motion to
solve problems
Free Falling
A tennis ball is thrown vertically upward with an initial
velocity of 8 m/s
What is the speed when ball return to its starting point
How long does the ball take to reach its starting point
𝑣" = 8
∆𝑦 = 0
𝑣!# = 𝑣"# − 2𝑔∆𝑦
𝑣! =? ?
𝑡 =? ?
𝑣! = 𝑣" − 𝑔𝑡
= 8# − 20(0)
−8 = 8 − 10𝑡
= 64
16
𝑡=
= 1.6 s
10
𝑣! = 64 = −8 m/s
After 1.6 s the
ball returns to its
starting point.
Therefore 0.8 s
required to reach
the maximum
height (peak).
use equations
of motion to
solve problems
Free Falling
A player throws volleyball upward with initial velocity of
11 m/s. if volleyball starts from 2 m above the floor.
What is the speed when it reaches the floor
How long does it take to reach the floor
𝑣" = 11
∆𝑦 = −2
𝑣!# = 𝑣"# − 2𝑔∆𝑦
𝑣! =? ?
𝑡 =? ?
𝑣! = 𝑣" − 𝑔𝑡
= 11# − 20(−2)
−12.7 = 11 − 10𝑡
= 121 + 40
23.7
𝑡=
= 2.37 s
10
𝑣! = 161 = −12.7 m/s
After 1 s the ball
will not reach the
maximum height
because the
curve does not
symmetrical.
explain free
falling graphs
Free Falling
From equations of motion we can draw position – time graph and
velocity – time graph.
∆𝑦(m)
𝑝𝑒𝑎𝑘
𝑣!
t(s)
𝑎(m/s # )
𝑣" (m/s)
𝑝𝑒𝑎𝑘
t(s)
t(s)
−𝑔
before peak upward
motion, after peak
downward motion
object is thrown.
before peak slowing
down, after peak
seeding up
here 𝑣! ≠ 0.
the line on negative
y – axis because the
motion is downward
and its constant.