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General
Physics 101
Freely Falling Objects
Prepared By
Prof. Rashad Badran
By Prof. Rashad Badran
Freely Falling Objects
• A freely falling Object is any object falling freely under the influence of
gravity, regardless of its initial motion.
• The ideal case of this motion occurs when the air resistance is
negligible.
• Thus, only one force acts on the object, that is, the force of gravity.
By Prof. Rashad Badran
One Dimensional Motion: An Application
Galileo
• Two different weights dropped
simultaneously from Tower of Pisa hit
the ground at approximately the same
time!!
• Rate of fall does not depend on mass. Leaning Tower of Pisa
Freely Falling Objects
By Prof. Rashad Badran
One Dimensional Motion:
An object may be released under the effect of gravity only in
three different cases:
(1)
From rest
(2)
Thrown downward
with an initial speed
(3)
Thrown upward
with an initial speed
vyi< 0 vyi> 0
vyi= 0
By
Prof.
Rashad
Badran
One Dimensional Motion:
Freely Falling Objects
(1) At the Earth’s surface, the free-fall acceleration (or
acceleration due to gravity) does not change with altitude
over short vertical distances: Freely falling object along y-axis
is approximately equivalent to the motion of a particle under
constant acceleration in 1-D.
ay= - g where g is acceleration due to gravity which has a
magnitude of 9.8 m/s2.
(2) Air resistance is neglected.
In the three different cases, two assumptions can be adopted:
By Prof. Rashad Badran
One Dimensional Motion:
Freely Falling Objects
t
y
a
yi
v
yf
v 

t
yf
v
yi
v
i
y
f
y )
(
2
1



2
2
1
t
y
a
t
yi
v
i
y
f
y 


)
(
2
2
2
i
y
f
y
y
a
yi
v
yf
v 

 0
g=9.8
m/s
2
Recalling the equations of motion in one dimension by replacing x by y to
indicate that the motion is along y-axis.
By Prof. Rashad Badran
One Dimensional Motion:
Freely Falling Objects
gt
yi
v
yf
v 

t
yf
v
yi
v
y )
(
2
1



2
2
1
gt
t
yi
v
y 


y
g
yi
v
yf
v 

 2
2
2
0
g=9.8
m/s
2
Insert ay = - g into the equations of motion in 1-D to get:
i
y
f
y
y 


where
The minus sign for the free-fall acceleration is
already inserted into the equations
By Prof. Rashad Badran
One Dimensional Motion:
Freely Falling Objects
gt
yi
v
yf
v 

t
yf
v
yi
v
y )
(
2
1



2
2
1
gt
t
yi
v
y 


y
g
yi
v
yf
v 

 2
2
2
g=9.8
m/s
2
i
y
f
y
y 


where
y=
+ve
v
y
=
+ve
If the object moves upward and is displaced
upward then its velocity (vy) and
displacement (y) must have positive signs
By Prof. Rashad Badran
One Dimensional Motion:
Freely Falling Objects
gt
yi
v
yf
v 

t
yf
v
yi
v
y )
(
2
1



2
2
1
gt
t
yi
v
y 


y
g
yi
v
yf
v 

 2
2
2
g=9.8
m/s
2
i
y
f
y
y 


where

y=
-
ve
v
y
=
-
ve
If the object moves downward and is
displaced downward then its velocity (vy) and
displacement (y) must have negative signs
By Prof. Rashad Badran
One Dimensional Motion:
Freely Falling Objects
A ball is thrown directly downward with an initial speed of 8 m/s from a height of 30 m.
(a) after what time interval does it strike the ground?(b) what is its final speed just
before hitting the ground? [Hint: Take g= 10 m/s2]
Example
g=
10
m/s
2
v
yi
=
-
8m/s
h=
30
m
By Prof. Rashad Badran
One Dimensional Motion:
By Prof. Rashad Badran
(a) Here initial time is taken ti = 0. Initial velocity has a negative sign (i.e. vyi = - 8 m/s).
Also the displacement must have a negative sign (i.e. y = - 30 m). Substitute g = 10
m/s2. To find tf use the equation:
g=
10
m/s
2
v
yi
=
-
8m/s

y=
-
30
m
ti = 0
tf = ?
vyf = ?
Solution:
2
)
2
10
(
)
8
(
30
2
2
1
t
t
gt
t
yi
v
y







Δ
s
t
t
or
t
77
.
1
10
77
.
25
8
10
77
.
25
8








By Prof. Rashad Badran
One Dimensional Motion:
a
ac
b
b
t
2
4
2 



0
2


 c
bt
at
0
30
8
2
5 

 t
t
10
)
30
)(
5
(
4
2
)
8
(
8 




t
s
m
yf
v
yf
v
y
g
yi
v
yf
v
/
77
.
25
)
30
)(
10
(
2
2
)
8
(
2
2
2
2











(b) To find the final speed use the information: [vyi = - 8 m/s, y = - 30 m, and g= 10
m/s2] and employ the equation:
Note: The final speed is 25.77 m/s
while the negative sign refers to
the direction of velocity
(downward).
g=
10
m/s
2
v
yi
=
-
8m/s

y=
-
30
m
ti = 0
tf = 1.77 s
vyf = ?
Solution:
By Prof. Rashad Badran
One Dimensional Motion:
A ball is released from rest from a height of 30 m. (a) after what time interval does it
strike the ground?(b) what is its final speed just before hitting the ground? (c) where
is the ball (with respect to the level of thrower (or the top of building)) after one
second of its release? [Hint: Take g= 10 m/s2].
Example
g=
10
m/s
2
vyi = 0
h=
30
m
By Prof. Rashad Badran
One Dimensional Motion:
(a) Here initial time is taken ti = 0. Initial velocity vyi = 0. The displacement must have
a negative sign (i.e. y = - 30 m). Substitute g = 10 m/s2. To find tf use the equation:
g=
10
m/s
2
vyi = 0

y=
-
30
m
ti = 0
tf = ?
vyf = ?
Solution:
s
t
t
t
gt
t
yi
v
y
45
.
2
30
2
5
2
)
2
10
(
0
30
2
2
1











The ball strikes the ground after the
time interval 2.45 s By Prof. Rashad Badran
One Dimensional Motion:
s
m
yf
v
yf
v
y
g
yi
v
yf
v
/
5
.
24
)
30
)(
10
(
2
0
2
2
2
2










(b) To find the final speed use the information: [vyi = 0, y = - 30 m, and g= 10 m/s2]
and employ the equation:
Note: The final speed is 24.5 m/s
while the negative sign refers to
the direction of velocity
(downward).
g=
10
m/s
2
vyi = 0

y=
-
30
m
ti = 0
tf = 2.45 s
vyf = ?
Solution:
By Prof. Rashad Badran
One Dimensional Motion:
(c) To find the location of ball after 1 s of its release: use the information: [vyi = 0, and
g= 10 m/s2] and employ the equation:
Note: The ball is 5 m below the top
of the building (or 25 m from the
ground).
g=
10
m/s
2
vyi = 0

y=
?
ti = 0
tf = 1 s
Solution:
h=
30
m
m
y
y
gt
t
yi
v
y
5
2
)
1
)(
2
10
(
0
2
2
1











By Prof. Rashad Badran
One Dimensional Motion:
A ball is thrown vertically upward with an initial speed of 10 m/s from the top of a
building of 30 m height. (a) Using tA = 0 as the time the ball leaves the thrower’s hand
at point A, determine the time at which the ball reaches its maximum height.
(b) find the maximum height of the ball. (c) Determine the velocity of the ball when it
returns to the height from which it was thrown. (d) Find the position and velocity of
the ball at t = 3 s. (e) what is the ball’s velocity just before it strikes the ground [Hint:
Take g = 10 m/s2]
Example
g=10
m/s
2
vyi = 10 m/s
h=
30
m
tA = 0 A
By
Prof.
Rashad
Badran
One Dimensional Motion:
By Prof. Rashad Badran
Solution:
g=10
m/s
2
vA = 10 m/s
h=
30
m
tA = 0 A C
B
D
E
tB = ?
vB = 0
(y)AB =?
(a)
B
gt
A
y
v
B
y
v 

s
B
t
B
t
1
10
10
0




The time it takes the ball to reach its maximum
height is 1s.
Note: It can be easily shown that the time it takes
the ball to reach point C from A is 2s.
Turning point
By Prof. Rashad Badran
One Dimensional Motion:
Solution:
m
B
A
y
B
A
y
5
)
(
)
)(
10
(
2
2
)
10
(
0







The maximum height with respect to point A is 5 m
(b)
g=10
m/s
2
vA = 10 m/s
h=
30
m
tA = 0 A C
B
D
E
tB = 1 s
vB = 0
(y)AB =?
tC = 2 s
Note: The result shows that the displacement is
positive as the ball moves from point A from B.
B
A
y
g
A
y
v
B
y
v



 )
(
2
2
2
Turning point
By Prof. Rashad Badran
One Dimensional Motion:
Solution:
(c)
At the level of thrower’s hand (point
A), the ball returns to the same
position
g=10
m/s
2
vA = 10 m/s
h=
30
m
tA = 0 A vC = ?
C
B
D
E
tB = 1 s
vB = 0
(y)AB = 5 m
tC = 2 s
By Prof. Rashad Badran
One Dimensional Motion:
Solution:
(c)
At the level of thrower’s hand (point
A), the ball returns to the same
position, In this case (y)AC = 0
s
m
A
y
v
C
y
v
A
y
v
C
y
v
C
A
y
g
A
y
v
C
y
v
/
10
2
2
)
(
2
2
2









Note: The positive sign is discarded
because the ball is moving downward
g=10
m/s
2
vA = 10 m/s
h=
30
m
tA = 0 A vC = ?
C
B
D
E
tB = 1 s
vB = 0
(y)AB = 5 m
tC = 2 s
By Prof. Rashad Badran
One Dimensional Motion:
Solution:
(d) The ball is launched at point A with velocity vA=
+10m/s until it reaches point D, when the time reads
tD = 3 s, and the ball is displaced downward from A
to D (i.e. (y)AD = yD-yA). One can use this
information in the following equation:
Note: The ball is 15 m below the top of
building (point A).
m
D
A
y
D
A
y
D
gt
D
t
A
y
v
D
A
y
15
)
(
2
)
3
(
2
10
)
3
)(
10
(
)
(
2
2
1
)
(













g=10
m/s
2
vA = 10 m/s
h=
30
m
tA = 0 A vC = -10 m/s
C
B
D
E
tB = 1 s
vB = 0
tD = 3 s
vD = ?
(y)AB = 5 m
tC = 2 s
By Prof. Rashad Badran
One Dimensional Motion:
Solution:
Using the same information as before and the
following equation, one can find the velocity at point D
Note: The velocity has a magnitude of
20m/s. The minus sign indicates the
direction of velocity (downward)
s
m
D
y
v
D
y
v
D
gt
A
y
v
D
y
v
/
20
)
3
(
10
10







g=10
m/s
2
vA = 10 m/s
h=
30
m
tA = 0 A vC = -10 m/s
C
B
D
E
tB = 1 s
vB = 0
tD = 3 s
vD = ?
(y)AB = 5 m
tC = 2 s
By Prof. Rashad Badran
One Dimensional Motion:
Solution:
Note: The velocity has a magnitude of
26.46m/s and its direction is downward as
the minus sign indicates.
s
m
E
y
v
E
y
v
E
A
y
g
A
y
v
E
y
v
/
46
.
26
)
30
)(
10
(
2
2
)
10
(
2
)
(
2
2
2









(e) To find the final velocity, use the information: [vyA
= +10 m/s, y = - 30 m, and g= 10 m/s2] and employ
the equation:
g=10
m/s
2
vA = 10 m/s
h=
30
m
tA = 0 A vC = -10 m/s
C
B
D
E
tB = 1 s
vB = 0
tD = 3 s
vD =- 20 m/s
vE =?
(y)AB = 5 m
tC = 2 s
By Prof. Rashad Badran
One Dimensional Motion:

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Part-6-1-D-Motion-Free Fall-Physics 101.pdf

  • 1. General Physics 101 Freely Falling Objects Prepared By Prof. Rashad Badran By Prof. Rashad Badran
  • 2. Freely Falling Objects • A freely falling Object is any object falling freely under the influence of gravity, regardless of its initial motion. • The ideal case of this motion occurs when the air resistance is negligible. • Thus, only one force acts on the object, that is, the force of gravity. By Prof. Rashad Badran One Dimensional Motion: An Application
  • 3. Galileo • Two different weights dropped simultaneously from Tower of Pisa hit the ground at approximately the same time!! • Rate of fall does not depend on mass. Leaning Tower of Pisa Freely Falling Objects By Prof. Rashad Badran One Dimensional Motion:
  • 4. An object may be released under the effect of gravity only in three different cases: (1) From rest (2) Thrown downward with an initial speed (3) Thrown upward with an initial speed vyi< 0 vyi> 0 vyi= 0 By Prof. Rashad Badran One Dimensional Motion: Freely Falling Objects
  • 5. (1) At the Earth’s surface, the free-fall acceleration (or acceleration due to gravity) does not change with altitude over short vertical distances: Freely falling object along y-axis is approximately equivalent to the motion of a particle under constant acceleration in 1-D. ay= - g where g is acceleration due to gravity which has a magnitude of 9.8 m/s2. (2) Air resistance is neglected. In the three different cases, two assumptions can be adopted: By Prof. Rashad Badran One Dimensional Motion: Freely Falling Objects
  • 6. t y a yi v yf v   t yf v yi v i y f y ) ( 2 1    2 2 1 t y a t yi v i y f y    ) ( 2 2 2 i y f y y a yi v yf v    0 g=9.8 m/s 2 Recalling the equations of motion in one dimension by replacing x by y to indicate that the motion is along y-axis. By Prof. Rashad Badran One Dimensional Motion: Freely Falling Objects
  • 7. gt yi v yf v   t yf v yi v y ) ( 2 1    2 2 1 gt t yi v y    y g yi v yf v    2 2 2 0 g=9.8 m/s 2 Insert ay = - g into the equations of motion in 1-D to get: i y f y y    where The minus sign for the free-fall acceleration is already inserted into the equations By Prof. Rashad Badran One Dimensional Motion: Freely Falling Objects
  • 8. gt yi v yf v   t yf v yi v y ) ( 2 1    2 2 1 gt t yi v y    y g yi v yf v    2 2 2 g=9.8 m/s 2 i y f y y    where y= +ve v y = +ve If the object moves upward and is displaced upward then its velocity (vy) and displacement (y) must have positive signs By Prof. Rashad Badran One Dimensional Motion: Freely Falling Objects
  • 9. gt yi v yf v   t yf v yi v y ) ( 2 1    2 2 1 gt t yi v y    y g yi v yf v    2 2 2 g=9.8 m/s 2 i y f y y    where  y= - ve v y = - ve If the object moves downward and is displaced downward then its velocity (vy) and displacement (y) must have negative signs By Prof. Rashad Badran One Dimensional Motion: Freely Falling Objects
  • 10. A ball is thrown directly downward with an initial speed of 8 m/s from a height of 30 m. (a) after what time interval does it strike the ground?(b) what is its final speed just before hitting the ground? [Hint: Take g= 10 m/s2] Example g= 10 m/s 2 v yi = - 8m/s h= 30 m By Prof. Rashad Badran One Dimensional Motion: By Prof. Rashad Badran
  • 11. (a) Here initial time is taken ti = 0. Initial velocity has a negative sign (i.e. vyi = - 8 m/s). Also the displacement must have a negative sign (i.e. y = - 30 m). Substitute g = 10 m/s2. To find tf use the equation: g= 10 m/s 2 v yi = - 8m/s  y= - 30 m ti = 0 tf = ? vyf = ? Solution: 2 ) 2 10 ( ) 8 ( 30 2 2 1 t t gt t yi v y        Δ s t t or t 77 . 1 10 77 . 25 8 10 77 . 25 8         By Prof. Rashad Badran One Dimensional Motion: a ac b b t 2 4 2     0 2    c bt at 0 30 8 2 5    t t 10 ) 30 )( 5 ( 4 2 ) 8 ( 8      t
  • 12. s m yf v yf v y g yi v yf v / 77 . 25 ) 30 )( 10 ( 2 2 ) 8 ( 2 2 2 2            (b) To find the final speed use the information: [vyi = - 8 m/s, y = - 30 m, and g= 10 m/s2] and employ the equation: Note: The final speed is 25.77 m/s while the negative sign refers to the direction of velocity (downward). g= 10 m/s 2 v yi = - 8m/s  y= - 30 m ti = 0 tf = 1.77 s vyf = ? Solution: By Prof. Rashad Badran One Dimensional Motion:
  • 13. A ball is released from rest from a height of 30 m. (a) after what time interval does it strike the ground?(b) what is its final speed just before hitting the ground? (c) where is the ball (with respect to the level of thrower (or the top of building)) after one second of its release? [Hint: Take g= 10 m/s2]. Example g= 10 m/s 2 vyi = 0 h= 30 m By Prof. Rashad Badran One Dimensional Motion:
  • 14. (a) Here initial time is taken ti = 0. Initial velocity vyi = 0. The displacement must have a negative sign (i.e. y = - 30 m). Substitute g = 10 m/s2. To find tf use the equation: g= 10 m/s 2 vyi = 0  y= - 30 m ti = 0 tf = ? vyf = ? Solution: s t t t gt t yi v y 45 . 2 30 2 5 2 ) 2 10 ( 0 30 2 2 1            The ball strikes the ground after the time interval 2.45 s By Prof. Rashad Badran One Dimensional Motion:
  • 15. s m yf v yf v y g yi v yf v / 5 . 24 ) 30 )( 10 ( 2 0 2 2 2 2           (b) To find the final speed use the information: [vyi = 0, y = - 30 m, and g= 10 m/s2] and employ the equation: Note: The final speed is 24.5 m/s while the negative sign refers to the direction of velocity (downward). g= 10 m/s 2 vyi = 0  y= - 30 m ti = 0 tf = 2.45 s vyf = ? Solution: By Prof. Rashad Badran One Dimensional Motion:
  • 16. (c) To find the location of ball after 1 s of its release: use the information: [vyi = 0, and g= 10 m/s2] and employ the equation: Note: The ball is 5 m below the top of the building (or 25 m from the ground). g= 10 m/s 2 vyi = 0  y= ? ti = 0 tf = 1 s Solution: h= 30 m m y y gt t yi v y 5 2 ) 1 )( 2 10 ( 0 2 2 1            By Prof. Rashad Badran One Dimensional Motion:
  • 17. A ball is thrown vertically upward with an initial speed of 10 m/s from the top of a building of 30 m height. (a) Using tA = 0 as the time the ball leaves the thrower’s hand at point A, determine the time at which the ball reaches its maximum height. (b) find the maximum height of the ball. (c) Determine the velocity of the ball when it returns to the height from which it was thrown. (d) Find the position and velocity of the ball at t = 3 s. (e) what is the ball’s velocity just before it strikes the ground [Hint: Take g = 10 m/s2] Example g=10 m/s 2 vyi = 10 m/s h= 30 m tA = 0 A By Prof. Rashad Badran One Dimensional Motion: By Prof. Rashad Badran
  • 18. Solution: g=10 m/s 2 vA = 10 m/s h= 30 m tA = 0 A C B D E tB = ? vB = 0 (y)AB =? (a) B gt A y v B y v   s B t B t 1 10 10 0     The time it takes the ball to reach its maximum height is 1s. Note: It can be easily shown that the time it takes the ball to reach point C from A is 2s. Turning point By Prof. Rashad Badran One Dimensional Motion:
  • 19. Solution: m B A y B A y 5 ) ( ) )( 10 ( 2 2 ) 10 ( 0        The maximum height with respect to point A is 5 m (b) g=10 m/s 2 vA = 10 m/s h= 30 m tA = 0 A C B D E tB = 1 s vB = 0 (y)AB =? tC = 2 s Note: The result shows that the displacement is positive as the ball moves from point A from B. B A y g A y v B y v     ) ( 2 2 2 Turning point By Prof. Rashad Badran One Dimensional Motion:
  • 20. Solution: (c) At the level of thrower’s hand (point A), the ball returns to the same position g=10 m/s 2 vA = 10 m/s h= 30 m tA = 0 A vC = ? C B D E tB = 1 s vB = 0 (y)AB = 5 m tC = 2 s By Prof. Rashad Badran One Dimensional Motion:
  • 21. Solution: (c) At the level of thrower’s hand (point A), the ball returns to the same position, In this case (y)AC = 0 s m A y v C y v A y v C y v C A y g A y v C y v / 10 2 2 ) ( 2 2 2          Note: The positive sign is discarded because the ball is moving downward g=10 m/s 2 vA = 10 m/s h= 30 m tA = 0 A vC = ? C B D E tB = 1 s vB = 0 (y)AB = 5 m tC = 2 s By Prof. Rashad Badran One Dimensional Motion:
  • 22. Solution: (d) The ball is launched at point A with velocity vA= +10m/s until it reaches point D, when the time reads tD = 3 s, and the ball is displaced downward from A to D (i.e. (y)AD = yD-yA). One can use this information in the following equation: Note: The ball is 15 m below the top of building (point A). m D A y D A y D gt D t A y v D A y 15 ) ( 2 ) 3 ( 2 10 ) 3 )( 10 ( ) ( 2 2 1 ) (              g=10 m/s 2 vA = 10 m/s h= 30 m tA = 0 A vC = -10 m/s C B D E tB = 1 s vB = 0 tD = 3 s vD = ? (y)AB = 5 m tC = 2 s By Prof. Rashad Badran One Dimensional Motion:
  • 23. Solution: Using the same information as before and the following equation, one can find the velocity at point D Note: The velocity has a magnitude of 20m/s. The minus sign indicates the direction of velocity (downward) s m D y v D y v D gt A y v D y v / 20 ) 3 ( 10 10        g=10 m/s 2 vA = 10 m/s h= 30 m tA = 0 A vC = -10 m/s C B D E tB = 1 s vB = 0 tD = 3 s vD = ? (y)AB = 5 m tC = 2 s By Prof. Rashad Badran One Dimensional Motion:
  • 24. Solution: Note: The velocity has a magnitude of 26.46m/s and its direction is downward as the minus sign indicates. s m E y v E y v E A y g A y v E y v / 46 . 26 ) 30 )( 10 ( 2 2 ) 10 ( 2 ) ( 2 2 2          (e) To find the final velocity, use the information: [vyA = +10 m/s, y = - 30 m, and g= 10 m/s2] and employ the equation: g=10 m/s 2 vA = 10 m/s h= 30 m tA = 0 A vC = -10 m/s C B D E tB = 1 s vB = 0 tD = 3 s vD =- 20 m/s vE =? (y)AB = 5 m tC = 2 s By Prof. Rashad Badran One Dimensional Motion: