How to solve this question please answer?

♣ his velocity is a vector v=|v|*(i*cos(θ°) +j*sin(θ°)), where |v|=3 km/h in calm water; river velocity vector is u=i*|u|, where |u|=5 km/h; x-axis being parallel with river flow;

♠ we must take into account only his swimming to the opposite side of the river, that is y-component of vector v;

♦ thus to cross the river will take time

t=(0.5 km)/ (|v|*sinθ) =(0.5/3)/sinθ hours =

= ((0.5/3)/sinθ)*60 minutes = 10/sinθ minutes;

That's not the answer I get. The answer should be the same regardless of the velocity of the water. The only thing affected by the water velocity is where he ends up on the other side.

If he swims perpendicular to the opposite bank, the time required will be: t = D / V

T = 0.5 km / 5 km/hr = 0.1 hr (a tenth of an hour is 6 min., not 10 minutes.) If he swims at angle theta to the opposite bank (or the flow), then the time required will be

(6 minutes)/ (sin theta)

1/2 km / 3 km/h = 1/6 h = 10 minutes

If he heads straight to the other shore, 90° to the flow he needs 10 minutes to cross:

Theta = 90°

sin 90° = 1

10 minutes / 1 = 10 minutes

If he would cross at an angle of 30° (either upstream or downstream) he would need 20 minutes to cross:

Theta = 30°

sin 30° = 0.5

10 minutes / 0.5 = 20 minutes

Anita Sood Swimmer

This Answer = (10/sin theta) minutes makes no sense !

What is theta ?

need more info

sarang it is not cricket , so better ask ANITA SOOD, ok.