(4)Projectile Motion

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  • Created on: 24-03-14 11:02
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W6.1 ­ Projectiles
This is where two dimensional motion is probably most easily done in
rather than in vectors.
We shall first consider an example taking place entirely on horizontal ground ­
projected from ground level and landing at ground level.
In all our examples, we shall ignore air resistance.
A ball is kicked at 15ms-1 at an angle of 25° above the horizontal.
How high does it go
and how far away does it land?
Its initial velocity can be split into two components:
15 cos 25° horizontally
and 15 sin 25° vertically upwards.
We shall first find how high it goes.
There are two ways.
Both of them depend on locating where or when the vertical component
of the velocity is zero.
First, to locate where that happens.
= + 2as 0= 2gs
s= = 2.05m
We could also find when that happens.
v = u + at 0 = 15 sin 25° gt
t= = 0.646s
We shall see later that the particle's trajectory is as symmetrical as we're
inclined to draw it.
CT Training 02/05/2010

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That means that the time from A to B is half the time from A to C.
So the time to C will be 1.29s
It only remains to see how far it goes horizontally in that time:
s = ut s = 15 cos 25° 1.29 = 17.6m
Notice the importance in these examples of stating which of the constant
acceleration formulae you're using
and which direction you're counting as positive.…read more

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Obviously this will give the same answer.
You could spoil it, though, by not making a clear decision as to which
direction to count as positive.
As we want the horizontal distance to its landing point, we now apply
this time to the horizontal component:
s = ut s = 50 2.47 = 124m (3 s.f.)
There are some other details which can come into more developed questions.
One has the projectile just missing the top of a tree.…read more

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We shall consider the general problem on horizontal ground:
a particle projected at speed U at and angle of above the horizontal.…read more

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The symmetry of quadratic curves therefore justifies the dependence we have
had in this topic on the symmetry of the trajectory.…read more


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