Projektielbeweging e

Natasia Gouws
Natasia GouwsPhysical Science Teacher
Projektielbeweging e
Gravitational Acceleration
 Acceleration that an object
experiences in the absence of
resistance forces like air
resistance
 Symbol: g
 9,8 m·s-2 downwards
Gravitational Acceleration
From Newton’s Law of Universal Gravitation:
Where M = Mass of planet (kg)
R = Radius of planet (m)
G = Universal Gravitational Constant
= 6,67 × 10−11
N ∙ 𝑚2
∙ 𝑘𝑔−2
Gravitational Acceleration
The
motion of an object in
the of air
resistance, when
gravitational
force is exerted on it.
Freefall:
Freefall:
Terminal Velocity:
 Air friction = Weight
 Acceleration = 0 m·s-2
 Velocity remains constant
𝐹 𝐷
𝐹𝑔
𝑥 𝑣𝑠. 𝑡
𝑣 𝑣𝑠. 𝑡
𝑎 𝑣𝑠. 𝑡
Graphs of Motion: Calculations
Gradient of
𝑥 − 𝑡 graph
gives 𝑣
Gradient of
𝑣 − 𝑡 graph
gives 𝑎
Area under
𝑎 − 𝑡 graph
gives 𝑣
Area under
𝑣 − 𝑡 graph
gives 𝑥
Equations of Motion: Symbols
𝒗𝒊
𝒗 𝒇
∆𝒙/∆𝒚
𝒂
∆𝒕
Initial Velocity
Final Velocity
Displacement
Acceleration
Time
m·s-1
m·s-1
m
m·s-2
s
Vector
Vector
Vector
Vector
Scalar
𝒗 𝒇 = 𝒗𝒊 + 𝒂∆𝒕
𝒗 𝒇
𝟐
= 𝒗𝒊
𝟐
+ 𝟐𝒂∆𝒚
Equations of Motion
∆𝒚 = 𝒗𝒊∆𝒕 +
𝟏
𝟐
𝒂∆𝒕 𝟐
∆𝒚 =
𝟏
𝟐
(𝒗 𝒇 + 𝒗𝒊)∆𝒕
Equations of Motion
1) Choose and indicate a positive
direction.
2) Write down what is given
3) Write down what is asked
4) Choose equation and perform
calculation
Equations of Motion: Analysis
An object falls from a certain
above the ground
𝒉 = ∆𝒚
𝒚
(m)
t (s)
𝒗𝒊 = 𝟎
𝒗 𝒇 = 𝒗 𝒎𝒂𝒙
+
𝒉 = ∆𝒚
𝒗𝒊 = 𝟎
𝒗 𝒇 = 𝒗 𝒎𝒂𝒙
𝑣
(m·s-1)
t (s)
+
An object falls from a certain
above the ground
Example 1
In an experiment which resembles the
one Gallileo did, a ball is dropped from
the top of a building with a height of 80
m. Calculate:
a) The time it takes for the ball to reach
the ground.
b) The velocity with which the ball
reaches the ground.
Example 1
During the same experiment, a second
ball is dropped from the same height
1,5 s after the first one.
c) Calculate the velocity with which this
ball must be thrown in order for it to
reach the ground at the same time as
the first one.
h
𝒚
(m)
t (s)
𝒗 = 𝟎
𝒗 = 𝒗 𝒎𝒂𝒙
∆𝒚 = 𝟎
+
An object is thrown upwards and
returns to the same height.
h
t (s)
𝒗 = 𝟎
𝒗 = 𝒗 𝒎𝒂𝒙
𝑣
(m·s-1)+
An object is thrown upwards and
returns to the same height.
𝒚
(m)
t (s)𝟏
𝟐 ∆𝒕
A closer look at the graphs
+𝒗 𝒎𝒂𝒙
−𝒗 𝒎𝒂𝒙
t (s)
𝑣
(m·s-1)
A closer look at the graphs
Example 2
A boy throws a ball vertically upwards
with a velocity of 20 m·s-1. Calculate:
a) The maximum height that the ball
reaches.
b) The time it takes the ball to return to
the boys hand.
𝒚
(m) t (s)
∆𝒚
𝒗 = 𝟎
𝒗 = 𝒗 𝒎𝒂𝒙
𝒗 = 𝒗𝒊
+
An object is thrown upwards from
a point above the ground
t (s)
𝒗 = 𝟎
𝒗 = 𝒗 𝒎𝒂𝒙
𝒗 = 𝒗𝒊
𝑣
(m·s-1)
+
An object is thrown upwards from
a point above the ground
𝒚
(m) t (s)
−∆𝒚
𝒉 𝒎𝒂𝒙
𝒉 𝒃𝒐 𝒈𝒓𝒐𝒏𝒅
A closer look at the graphs
t (s)
𝑣
(m·s-1)
+𝒗𝒊
−𝒗𝒊
𝒗 𝒎𝒂𝒙
𝒉 𝒎𝒂𝒙
A closer look at the graphs
Example 3
A hot air balloon rises with a constant
velocity of 5 m·s-1. At a height of 60 m
above the ground, a sandbag is allowed
to drop. Assume that the balloon keeps
on moving with the same velocity.
Calculate:
a) The maximum height above the
ground that the bag will reach.
b) The distance between the sandbag
and the balloon at 3 s.
c) The time it takes the sandbag to reach
the ground.
d) The velocity with which the sandbag
reaches the ground.
Example 3
Example 3
𝒚
(m)
t (s)
t (s)
𝑣
(m·s-1)
Example 4
𝒗
(𝒎 ∙ 𝒔−𝟏
)
2,45
-2,45
-4,90
𝒕 (𝒔)
The above velocity-time graph describes
the motion of a bouncing ball that is
allowed to drop from a height of 1,23 m.
Choose downward negative for your
calculations.
Example 4
a) The skew downward lines are parallel.
Why?
b) How many times did the ball boumce on
the surface?
c) With what velocity does the ball reach the
ground the first time?
d) With what velocity does the ball leave the
ground the first time?
Example 4
e) How long did it take the ball to reach the
ground the first time after being dropped?
f) Show that the ball reaches a maximum
height of 0,31 m after the first bounce.
g) Draw a free-hand displacement-time
graph for the motion of the ball untill it
bounces the 2nd time.
Example 4
1 de 34

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Projektielbeweging e

  • 3.  Acceleration that an object experiences in the absence of resistance forces like air resistance  Symbol: g  9,8 m·s-2 downwards Gravitational Acceleration
  • 4. From Newton’s Law of Universal Gravitation: Where M = Mass of planet (kg) R = Radius of planet (m) G = Universal Gravitational Constant = 6,67 × 10−11 N ∙ 𝑚2 ∙ 𝑘𝑔−2 Gravitational Acceleration
  • 5. The motion of an object in the of air resistance, when gravitational force is exerted on it. Freefall:
  • 7. Terminal Velocity:  Air friction = Weight  Acceleration = 0 m·s-2  Velocity remains constant 𝐹 𝐷 𝐹𝑔
  • 8. 𝑥 𝑣𝑠. 𝑡 𝑣 𝑣𝑠. 𝑡 𝑎 𝑣𝑠. 𝑡 Graphs of Motion: Calculations Gradient of 𝑥 − 𝑡 graph gives 𝑣 Gradient of 𝑣 − 𝑡 graph gives 𝑎 Area under 𝑎 − 𝑡 graph gives 𝑣 Area under 𝑣 − 𝑡 graph gives 𝑥
  • 9. Equations of Motion: Symbols 𝒗𝒊 𝒗 𝒇 ∆𝒙/∆𝒚 𝒂 ∆𝒕 Initial Velocity Final Velocity Displacement Acceleration Time m·s-1 m·s-1 m m·s-2 s Vector Vector Vector Vector Scalar
  • 10. 𝒗 𝒇 = 𝒗𝒊 + 𝒂∆𝒕 𝒗 𝒇 𝟐 = 𝒗𝒊 𝟐 + 𝟐𝒂∆𝒚 Equations of Motion
  • 11. ∆𝒚 = 𝒗𝒊∆𝒕 + 𝟏 𝟐 𝒂∆𝒕 𝟐 ∆𝒚 = 𝟏 𝟐 (𝒗 𝒇 + 𝒗𝒊)∆𝒕 Equations of Motion
  • 12. 1) Choose and indicate a positive direction. 2) Write down what is given 3) Write down what is asked 4) Choose equation and perform calculation Equations of Motion: Analysis
  • 13. An object falls from a certain above the ground 𝒉 = ∆𝒚 𝒚 (m) t (s) 𝒗𝒊 = 𝟎 𝒗 𝒇 = 𝒗 𝒎𝒂𝒙 +
  • 14. 𝒉 = ∆𝒚 𝒗𝒊 = 𝟎 𝒗 𝒇 = 𝒗 𝒎𝒂𝒙 𝑣 (m·s-1) t (s) + An object falls from a certain above the ground
  • 15. Example 1 In an experiment which resembles the one Gallileo did, a ball is dropped from the top of a building with a height of 80 m. Calculate: a) The time it takes for the ball to reach the ground. b) The velocity with which the ball reaches the ground.
  • 16. Example 1 During the same experiment, a second ball is dropped from the same height 1,5 s after the first one. c) Calculate the velocity with which this ball must be thrown in order for it to reach the ground at the same time as the first one.
  • 17. h 𝒚 (m) t (s) 𝒗 = 𝟎 𝒗 = 𝒗 𝒎𝒂𝒙 ∆𝒚 = 𝟎 + An object is thrown upwards and returns to the same height.
  • 18. h t (s) 𝒗 = 𝟎 𝒗 = 𝒗 𝒎𝒂𝒙 𝑣 (m·s-1)+ An object is thrown upwards and returns to the same height.
  • 19. 𝒚 (m) t (s)𝟏 𝟐 ∆𝒕 A closer look at the graphs
  • 20. +𝒗 𝒎𝒂𝒙 −𝒗 𝒎𝒂𝒙 t (s) 𝑣 (m·s-1) A closer look at the graphs
  • 21. Example 2 A boy throws a ball vertically upwards with a velocity of 20 m·s-1. Calculate: a) The maximum height that the ball reaches. b) The time it takes the ball to return to the boys hand.
  • 22. 𝒚 (m) t (s) ∆𝒚 𝒗 = 𝟎 𝒗 = 𝒗 𝒎𝒂𝒙 𝒗 = 𝒗𝒊 + An object is thrown upwards from a point above the ground
  • 23. t (s) 𝒗 = 𝟎 𝒗 = 𝒗 𝒎𝒂𝒙 𝒗 = 𝒗𝒊 𝑣 (m·s-1) + An object is thrown upwards from a point above the ground
  • 24. 𝒚 (m) t (s) −∆𝒚 𝒉 𝒎𝒂𝒙 𝒉 𝒃𝒐 𝒈𝒓𝒐𝒏𝒅 A closer look at the graphs
  • 25. t (s) 𝑣 (m·s-1) +𝒗𝒊 −𝒗𝒊 𝒗 𝒎𝒂𝒙 𝒉 𝒎𝒂𝒙 A closer look at the graphs
  • 26. Example 3 A hot air balloon rises with a constant velocity of 5 m·s-1. At a height of 60 m above the ground, a sandbag is allowed to drop. Assume that the balloon keeps on moving with the same velocity. Calculate: a) The maximum height above the ground that the bag will reach.
  • 27. b) The distance between the sandbag and the balloon at 3 s. c) The time it takes the sandbag to reach the ground. d) The velocity with which the sandbag reaches the ground. Example 3
  • 31. Example 4 𝒗 (𝒎 ∙ 𝒔−𝟏 ) 2,45 -2,45 -4,90 𝒕 (𝒔)
  • 32. The above velocity-time graph describes the motion of a bouncing ball that is allowed to drop from a height of 1,23 m. Choose downward negative for your calculations. Example 4
  • 33. a) The skew downward lines are parallel. Why? b) How many times did the ball boumce on the surface? c) With what velocity does the ball reach the ground the first time? d) With what velocity does the ball leave the ground the first time? Example 4
  • 34. e) How long did it take the ball to reach the ground the first time after being dropped? f) Show that the ball reaches a maximum height of 0,31 m after the first bounce. g) Draw a free-hand displacement-time graph for the motion of the ball untill it bounces the 2nd time. Example 4