projectile motion amazing shit will get you a 7 in ib
TRANSCRIPT
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7/29/2019 Projectile Motion Amazing Shit will get you a 7 in IB.
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Fields and Forces
Topic 9.1 Projectile Motion
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Components of Motion
When a body is in free motion, (movingthrough the air without any forces apart from
gravity and air resistance), it is called aprojectile
Normally air resistance is ignored so the onlyforce acting on the object is the force due to
gravityThis is a uniform force acting downwards
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Therefore if the motion of the projectile isresolved into the verticaland horizontal
componentsThe horizontal componentwill be unaffectedas there are no forces acting on it
The vertical componentwill be accelerated
downwards by the force due to gravity
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These two components can beconsidered as independent factors in
the motion of a projectile in a uniformfield
In the absence of air resistance the
path taken by anyprojectile is parabolic
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Solving Problems
In solving problems it is necessary toconsider the 2 components
independently
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Therefore the horizontal motion it isnecessary to use the equation
speed = distancetime
Where speed is the horizontal
component of the velocity
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Therefore the vertical motion it isnecessary to use the kinematic
equations for uniform accelerationi.e. Using the s.u.v.a.tequations
Where uand vare the initialand final
verticalcomponents of the velocity
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To be able to calculate the horizontal
distance we need to know thehorizontal speed, and the time.
The horizontal distance is easy tocalculate by resolving the velocity
40.0o
10.0ms-110.0 sin 40.0o
10.0 cos 40.0o
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However, to calculate the time we willneed to use the vertical component and
the s.u.v.a.t. Equations
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s= ?
u= 10.0 sin 40.0o ms-1
v= ?a= -10.0 ms-2 (Up is positive, thereforeacceleration here is negative)
t= ?We only have 2 of the values when weneed three to find any other
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However, if we ignore air resistance,then the final vertical component of the
velocity will be equal and opposite ofthe initial component
i.e. v= -10.0 sin 40.0o ms-1
Looking at the equations for uniformacceleration, we need an equation thatlinks u, v, aand t.
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v = u + at
Rearranging to make tthe subject
t = v
ua
Substitute in
t =-10.0 sin 40.0o
10.0 sin 40.0o
-10
t= 1.286 seconds
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Now returning to the horizontalcomponents
Using speed = distancetime
Rearranging distance = speed x time
Distance = 10.0 cos 40.0o x 1.286
Distance = 9.851 = 9.9 metres
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Using the Conservation ofEnergy
In some situations the use of theconservation of energy can be a muchsimpler method than using the kinematicequations
Solving projectile motion problems makes useof the fact that Ek+ Ep=constant at every
point in the objects flight (assuming no lossof energy due to friction)
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Example
A ball is projected at 25.0 ms-1 at anangle of 40.00 to the horizontal. The ball
is released 2.00m above the ground.Taking g = 10.0 ms-2. Find themaximum height it reaches.
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Solution
2.0m
25.0 ms-1
A
Bv = v
horizontal
H
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Total energy at A is given by
Ek+ Ep= m (25.0)2+ mg x 2.0
=312.5m + 20m
= 332.5m
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Next, to find the total energy at B weneed to know the velocity at B, which is
given by the horizontal component ofthe velocity at A
Total energy at B is given by
Ek
+ Ep
= m (25.0 cos 40o)2+ mg x H
=183.38m+ 10mH
Then using the conservation of energy
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Equating the 2 equations
332.5m = 183.38m + 10mH
332.5 = 183.38 + 10H
332.5183.38 = 10H
10H = 149.12
H = 14.912 = 14.9m