Question

a person, reaches a point directly opposite on the other bank of a flowing river while swimming at a speed of 5m/s at an angle of 120 degree with the flow. the speed of the flow must be

Answer 1

Let the velocity of of  man be Vm and velocity of  flow of  water Vw as  shown in figure.

First  method:-

If  we  divide Vm in two components as  it  is  a vector  quantity, Then,

Horizontal component Vm.cos120° will be  equal to Vw,

Vm.cos120° = Vw

Vw =  Vm.cos120°

Vw = - 5 × 1/2

Vw = - 2.5

|Vw| = 2.5 m/s

Second  Method,

In ΔABC,

sin30° = AB/AC

1/2 = Vw/Vm

Vw = 5 × 1/2

Vw = 2.5 m//s

Answer 2

"The speed of water is 2.5 m/s.

Solution:

${ V }_{ w }$ = Velocity of water

${ V }_{ m }$ = Velocity of man

Consider the triangle ABC,

In that, $\sin 30\quad =\quad \frac { AB }{ AD }$

$\frac { 1 }{ 2 } \quad =\quad \frac { { V }_{ { w } } }{ { V }_{ { m } } }$

Where ${ V }_{ w }$ is the velocity of the flow of water and ${ V }_{ m }$ is the velocity of the flow of man.

$V_{ { w } }\quad =\quad 5\quad \times \quad \frac { 1 }{ 2 }$

${ V }_{ { w } }\quad =\quad 2.5\quad { m }/{ s }$"