IIT JEE , AIEEE Forum - Physics , Mechanics

kghedriu

A launch travels across the river from a point A to point B on the opposite bank along the straight line AB forming an angle

as shown in the figure.The flag on the mast of the launch makes an angle

with the direction of its motion.Determine the speed of the launch with respect to the bank. Let 'u' be the velocity of the wind blowing perpendicularly to the current.

the answer given is u sin[

+

-

/2] sin

but i m getting the answer as u sin[

+

-

/2] cosec

i want to cross check my result...pls reply soon....

kghedriu

Sir...here is my solution.....if there are any mistakes, please point them out bcos the answer is not matching...

Vag => vel of air wrt ground

Vbg=> vel of boat wrt ground

Vwg => vel of water wrt ground

Vab => vel of air wrt boat

we know that the angle made by flag on the mast is due to flow of air wrt motion of boat....hence frm the fig, applying sine rule, we hav..

Vbg/ sin[alpha+beta-pi/2] = u/sin[pi-beta]

so...

Vbg = u sin[alpha+beta-pi/2] cosec(beta)

but the answer mentioned is having sin(beta) in place of cosec(beta)

i donno exactly whether ma method is right or not...

so..pls help me..

......................................Goutham..........................

THIS IS ALSO MY 400TH POST!!

(in just one and half month

)

pinnaka

goutham sorry for the delayed reply: by the way it is a nice question
Can u tell me the source of the problem in mail
Your method is perfectly correct the answer is given by your answer
I think we have to find the velocityof the lauch with respect to the river.

Here is my method of doing the problem

M = Velocity of the wind relative to launch
Since we know that the flag mast must be in the direction of M
Therefore the flagmast is in the direction of M
Now lets us take v to be respect to the ground or the bank
Then u = M+v
Now we will modify your diagram:
replace the ray v_ab with M and v_ag = u and the ray v_bg with v
Now look at your figure:
the angle that M makes with u is :aplha+beta-pi/2
and the other angle is pi-beta and not pi/2-beta (think about it, My statements are perfectly correct)
so now according to the sine theroem:
we have
v/sin[alpha+beta_pi/2) = u/(sin(pi-beta)
therfore u get v =u*sin( aplha+beta-pi/2)/sin(pi-beta)
manuplating the answer u v =u*sin( aplha+beta-pi/2)/sin(beta)
So we agree on a common answer
the source answer is wrong
--Stuart Anderson

edison

Your approach is Perfect Goutham Harsha.

Well done keep it up.

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