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Physics. Momentum in two dimensions.

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student:
That's a question from A2 ActiveBook.
http://img717.imageshack.us/img717/7660/fffvm.jpg

There are no answers, so I'm not sure if it's correct..
I assume it's this way. But I dont understand which side does he jump off.
http://img685.imageshack.us/img685/5949/p1100041ss.jpg

If he jumps in eastern direction:
a)
pN = 100*12*sin80 = 1181.77 kgms-1
pE = 200*3+100*12*cos80 = 808.38 kgms-1
v = (sqrt(1181.772+808.382)/300)=4.77 ms-1
angle between this v and the stream = tan-1(1181.77/808.38)=55.6°

b)
vN = 1181.77/300 = 3.94 ms-1
time to reach the bank = 16/3.94 = 4.06 s
vE = 808.38/300 = 2.69 ms-1
time to reach the waterfall = 100/2.69 = 37.11 s
4.06 < 37.11
So he will reach the bank before he falls off.

 ???

nid404:
I can't see the question

holtadit:

--- Quote from: nid404 on April 22, 2010, 06:52:56 am ---I can't see the question

--- End quote ---

Right click and select view image

student:
I'm sorry..
Posted the links to the images.

nid404:
I'm not too sure either...
But I'll still post mine

Check the diagram....

the velocity in direction of current will now have an component of 12cos 80o
So it will flow down the stream with a velocity of= 3 + 12cos80=5.08m/s
now i don't knw whether momentum is conserved...if so there is going to be some change
100X12cos80 + 3X200= 300 v  [using law of conservation of momentum]
this gives v= 2.7 m/s  :-\


I'm still in AS student...so  :(
anyway i gave it a try...astar will be able to answer this for you

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