Projectile Motion Amazing Shit will get you a 7 in IB

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

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Projectile Motion Amazing Shit will get you a 7 in IB