Physics. Momentum in two dimensions.

[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)
p_N = 100*12*sin80 = 1181.77 kgms^-1
p_E = 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)
v_N = 1181.77/300 = 3.94 ms^-1
time to reach the bank = 16/3.94 = 4.06 s
v_E = 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]

I can't see the question

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 80^o
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

[astarmathsandphysics]

my solutions

[student]

Thanks, i think it's rigth then, as the velocity i've got is the same as astar's.
Thanks nid404 for tying to help.