Embed Size (px)
Citation preview
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
𝐹𝐷
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
𝒗𝒊𝒗𝒇
∆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
𝒗𝒇 = 𝒗𝒊 + 𝒂∆𝒕
𝒗𝒇𝟐= 𝒗𝒊
𝟐+ 𝟐𝒂∆𝒚
Equations of Motion
∆𝒚 = 𝒗𝒊∆𝒕 +𝟏
𝟐𝒂∆𝒕𝟐
∆𝒚 =𝟏
𝟐(𝒗𝒇 + 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 = ∆𝒚𝒚(m)
t (s)
𝒗𝒊 = 𝟎
𝒗𝒇 = 𝒗𝒎𝒂𝒙
+
𝒉 = ∆𝒚
𝒗𝒊 = 𝟎
𝒗𝒇 = 𝒗𝒎𝒂𝒙
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.
h𝒚(m)
t (s)
𝒗 = 𝟎
𝒗 = 𝒗𝒎𝒂𝒙
∆𝒚 = 𝟎
+
An object is thrown upwards and
returns to the same height.
h t (s)
𝒗 = 𝟎
𝒗 = 𝒗𝒎𝒂𝒙
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
+𝒗𝒎𝒂𝒙
−𝒗𝒎𝒂𝒙
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 = 𝒗𝒎𝒂𝒙
𝒗 = 𝒗𝒊
+
An object is thrown upwards from
a point above the ground
t (s)
𝒗 = 𝟎
𝒗 = 𝒗𝒎𝒂𝒙
𝒗 = 𝒗𝒊
𝑣(m·s-1)
+
An object is thrown upwards from
a point above the ground
𝒚(m) t (s)
−∆𝒚
𝒉𝒎𝒂𝒙
𝒉𝒃𝒐 𝒈𝒓𝒐𝒏𝒅
A closer look at the graphs
t (s)
𝑣(m·s-1)
+𝒗𝒊
−𝒗𝒊
𝒗𝒎𝒂𝒙
𝒉𝒎𝒂𝒙
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
