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chetan
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#1
Report Thread starter 16 years ago
#1
can any1 help me out on an edexcel M2 question? if u got the M2 book its review exercise 1 page83 question 4c

i dnt understand how 2 calculate the angle???
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#2
Report 16 years ago
#2
Find the vertical component of the velocity:

v^2 = u^2 + 2as
v = root[(-10 x 3/5)^2 + 2 x 9.8 x 26.1]
v = 23.4

The horizontal component of the velocity is constant, ie 10cos alpha = 8

tan beta = 23.4/8
beta = 71 degrees.

Draw a triangle to help you see that, if it helps. The vertical component on the opposite side and the horizontal on the adjacent side. Hope that makes sense.
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chetan
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#3
Report Thread starter 16 years ago
#3
o i c ye i get it now thanx a lot
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#4
Report 16 years ago
#4
ok i just did it

i assume you got a) and b) correct?? Anyway at A the particle has horizontal velociry of 3x10 [email protected] = 8m/s, and vertical velocity of 10 [email protected] - 3g = 23.4m/s

Now the tangent of the angle created by the direction of motion is 23.4 / 8 = 2.925. So the angle is arctan 2.925 which is 71 degrees, the answer in the back.
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bloodhound
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#5
Report 16 years ago
#5
i was just doing it and someone else already posted it damn it.
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#6
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#6
No prob.
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#7
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#7
Originally posted by bloodhound
i was just doing it and someone else already posted it damn it.
Hehe, sorry .
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