Post on 04-Apr-2018
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Equations of ConstantAcceleration
Applications of Newtons Laws of
Motion
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Objectives
Understand projectile motion and howgravity influences a projectile.
Understand the effects of projectile speed,relative height and projection angle onprojectile motion.
Learn to compute maximum height, flighttime and range using equations of uniformacceleration.
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Projectile
A projectile is a body or object that is in the
air.
Only under the influence of gravity and air
resistance.
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Gravity
Gravity influences a projectile by pulling
the object toward the earth.
Acceleration due to gravity = - 9.8 m.s-1
Gravity causes a projectile to follow a
parabolic path.
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Why are projectiles important?
Success of many sporting events involvesprojectiles
Objects acting as projectiles Basketball, discus, others?
People acting as projectiles
Variables of interest Flight distance (range)
Flight time
Maximum height
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Equations of Constant
Acceleration
d = vit + .a.t
vf= vi + a.t
vf = vi + 2.a.d
Where
d = displacement
vi. = initial velocity
vf= final velocity
a = acceleration
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Equations of Constant
Acceleration
The equations are most commonly applied
when a rigid body is in flight and air
resistance is ignored and therefore the onlyforce acting is gravity.
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Applications of Equations
If a trampolinist is in the air for two
seconds, how high does he/she jump?
Questions: Which equation might you use?
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Applications of Equations
What is the velocity of the athlete at the
very top of the activity?
Answer: vi = 0
What is the value for gravity?
Answer: g = -9.8 m.s.-2
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Applications of Equations
Therefore:
d = vi
.t + .a.t
d = (0.0)(1) + (-9.8)(1)
d = -4.905 m
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Application II
A figure skater is attempting a jump in which she
performs 3 revolutions while in the air. She leaves
the ice with a velocity of 7 m/s at a projectionangle of 30. How long will she be in the air?
Is she spins 3 revolutions per second, will she be
able to complete all 3 revolutions before landing?
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Applications of Equations II
A ball is kicked from the ground level at an
angle of 60 degrees to the horizontal at a
velocity of 20 m.s 1.
Ignoring air resistance, how far does it
travel?
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Applications of Equations II
Resolve velocity vector into horizontal and
vertical components
Therefore, velocity in x direction is 10 m.s-1
Velocity in the y direction is 17.3 m.s-1
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Applications of Equations II
Time of flight is determined by the velocity
in the up direction (y direction)
Which equation of motion could we use to
determine the time of flight?
Answer: vf= vi + a.t
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Applications of Equations II
vi = 17.3 m.s-1
vf
= 0.0 m.s-1
a = -9.8 m.s-2
tup =1.76 s
tflight = 3.52 s
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Application of Equations II
What is the distance traveled horizontally?
What equation could we use?
Answer: d = vi.t + .a.t
There is no force acting
horizontally,therefore a = 0.0
d = 35.2 m
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Application of Equations II
Conclusion
The distance traveled depends on the time
in flight which depends on the vertical
velocity and horizontal velocity.
For a maximum distance to be obtained, a
high vertical and high horizontal velocity is
required
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Application of Equations II
The best compromise between the two
when resolving the velocity is at an angle of
45 degrees.
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Application of Equations III
Throwing
An athlete throwing from a projectile does
not release from ground height.
Therefore, there is a more complex
relationship between optimum angle of
release and height of release.
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Application of Equations III
The relationship is described by the
following equation.
R = (V/g).cos().
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Application of Equations III
As a general rule, the higher the release the
further the projectile goes.
As height of release increases, the angle of
release decreases.
For a shot putter releasing at a height of
2.44 m, the optimum angle is 40 degrees.
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Influences on Projectiles
Projection speed
Projection angle
Relative height