Transcript Vertical Projectiles

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Vertical projectile motion deals with objects
that fall straight down, objects that get
thrown straight up and the motion of an
object as it goes straight up and then down.
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The acceleration of this object due to gravity
is called gravitational acceleration, g, and is
equal to 9,8 m•s-2 on earth.
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Consider an object dropped from the top of a
building.
It accelerates at 9,8 m•s-2 and this is always
downwards.
The object has an initial velocity, vi of 0 m•s1.
The displacement, Δy, of the object is equal
to the height from which it falls.
The object reaches maximum velocity, vf, on
impact with the ground.
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Consider an object that is thrown vertically
upward.
As it get higher and higher, it slows down
until it stops momentarily at its highest point.
It has an initial velocity, vi, with which it was
projected i.e. vi is not 0 m•s-1
The acceleration is 9,8 m•s-2 downward.
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At the highest point the object stops and
therefore its final velocity vf = 0 m•s-1, but
the acceleration is still 9,8 m•s-2.
The object speeds up as it ascends.
The time taken for an object to reach its
maximum height is the same as the time it
takes to come back.
1. One word/term items
1.1. The force that acts on a body in free fall.
1.2. Motion of an object near the surface of the
earth under the influence of the earth’s
gravitational force alone.
2. Multiple Choice
2.1 Which of the following is a correct
statement?
Gravitational force is
A. applicable only in our solar system
B. both an attractive and repulsive force
C. directly proportional to the product of the
masses involved.
D. Directly proportional to both the masses
and the radius of the earth.
2.2 A golf ball is hit vertically upwards. What is
the acceleration of the ball at the highest
point? Ignore the effects of friction.
A. 9,8 m•s-2 upwards
B. 9,8 m•s-2 downwards
C. 0 m•s-2
D. 6,8 m•s-2 downwards
Long question
3. A stone is thrown vertically upwards at an
initial velocity of 24 m•s-1. It reaches its
maximum height after 2s.
3.1 Describe the motion of the stone in terms
of velocity and acceleration.
3.2 How long from the time it is thrown
upward, will it take to come back into the
thrower’s hand?
5.3 What
turning
5.4 What
turning
is the velocity of the stone at the
point?
is the acceleration of the stone at the
point?
1.1 Gravitational force/weight
1.2. Free fall
21. C
2.2. B
3.1 Initially the stone has a velocity of 24 m•s1 upwards. Throughout the motion it
experiences a constant downward
acceleration of 9,8m•s-2. its velocity
decreases to zero at the turning point. There
its direction of motion changes and the
velocity increases while moving downwards.
3.2 4s
3.3 0 m•s-1
3.4 9,8 m•s-2
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Projectiles can have their own motion
described by a single set of equations for the
upward and downward motion.
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At any point during the journey the
acceleration of the object is equal to the
gravitational acceleration, g.
g is equal to 9,8 m•s-2 downwards.
g is independent of the mass of an object.
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vf = vi + gΔt
Δy = viΔt + ½ gΔt2
vf2= vi2 + 2gΔy
Δy = (vf + vi)•Δt
2
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Choose a direction as positive.
Write down the values of the known vf; vi; g;
Δy and Δt
If an object is released or dropped by a
person that is moving up or down at a certain
velocity the initial velocity of an object equals
the velocity of that person.
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Identify which formula to use.
Substitute into the equation.
Interpret the answer.
1. A ball is thrown vertically upwards and
returns to the thrower’s hand 4s later.
Calculate
a) the height reached by the ball
b) the velocity with which the ball left the
thrower’s hand.
c) the velocity with which the ball returned to
the thrower’s hand.
2. A grade 12 learner wants to determine the
height of the school building. He projects a
stone vertically upward so that it reaches the
top of a building. The stone leaves a learner’s
hand at a height of 1,25m above the ground,
and he catches the stone again at a height of
1,25m above the ground. He finds the total
time the stone is in the air is 2s.Calculate the
height of the school building if air resistance
is ignored.
1.
a) Δy = viΔt + ½ gΔt2
= 0 (2) + ½ (9,8)(2)2
= 19,6 m
b)
vf2 = vi2 + 2gΔy
02 = vi2 + 2(-9,8)(19,6)
vi =19,6 m•s-1
2.
Take the upward motion as positive
vf = vi + gΔt
0 = vi + (-9,8)(1)
vi = 9,8 m•s-1 upward
Δy = viΔt + ½ gΔt2
= (9,8)(1) + ½ (-9,8)(1)2
= 4,9 m
Therefore total height = 6,15 m
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Equations of motion are equations that are
used to determine the motion of a body while
experiencing a force as a function of time.
These equations apply only to bodies moving
in one dimension/straight line with a
constant acceleration.
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The body’s motion is considered between two
time points: that is, from one initial point and
its final point in time.
Motion can be described in different ways:
◦ Words
◦ Diagrams
◦ Graphs
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We use three different graphs
◦ velocity – time graph
◦ acceleration – time graph
◦ position – time graph
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Consider a basketball player throwing a ball
in the air. What goes up must come down.
The ball has downward force acting on it
because of gravity. Therefore it will slow
down at a rate of 9,8 m∙s-1 after every
second. So we can say that the acceleration
is -9,8m∙s-2.
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When we tackle problems like this, we use the
equations of motion.
We have to make sure that we get the signs
right. We will make upwards positive and
downwards negative.
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Consider a ball thrown vertically upwards and
it returns to thrower’s hand.
We can represent these motions
graphically. It is important that you
understand these graphs.
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An object is dropped from a hot air balloon
which is ascending at a constant speed of
2m•s-1. Ignore the effects of air resistance.
a) Calculate how far below the point of
release the object will be after 4s.
b) Draw velocity vs time and acceleration vs
time graphs for the motion of the object from
the moment it is dropped from the balloon
until it hits the ground.
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a) Take up as positive
g =- 9,8 ms-1
vi = 2 ms-1
Δt = 4s
Δy = ?
Δy = viΔt + ½ gΔt2
= (2)(4) + ½ (-9,8)(4)2
= 70,4 m