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W6.1 ­ Projectiles
This is where two dimensional motion is probably most easily done in
components
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…

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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…

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s = ut + 30 = 0 + 0.5g
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…

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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.

First, to find the time to its highest point:
v = u + at 0 = U sin gt


t=
Twice that time is how long it…

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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.


Drill
Ex 7b (cont)
HW
HWR4 ­ Projectiles
Test
D2 Friction





CT Training 02/05/2010

Comments

daviesg

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Well presented walk through the worked examples on projectiles.  Explanation with the method. 

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