Question

a man swims across a river and reaches a point directly opposite in time t1 he swims an equal distance downstream and covers distance in time t2 if speed of man in still water is v and speed of river is u the t1/t2

Answer 1

Final Answer :
$\frac{t _{1}}{t _{2} } = \sqrt{ \frac{u + v}{u - v} }$
Steps:
1) Case -1 : Let the man starts by making an angle $\theta$ with vertical.
Velocity of man wrt river = v
Velocity of river = u.

Since, Man reaches directly opposite to river bank.
=> Horizontal velocity of man with respect to ground is 0.
$=> u - v\sin(\theta) = 0 \\ => u = v \sin(\theta) \\ => \cos(\theta) = \sqrt{1-\frac{u^2}{v^2}}$

2) Time taken by man to reach,eactly at oppose bank ,
$t_1 = \frac{d}{v\cos(\theta)} \\ t_1 = \frac{d}{\sqrt{v^2-u^2}}$

3) Case : 2 :(Downstream)
Velocity of man wrt ground along downstream : v + u
Distance Covered : d
$t_2 = \frac{d}{v+u}$

We get ,
$\frac{t_1}{t_2} = \sqrt{\frac{u+v}{u-v}}$

Answer 2

Answer: $t_1 / t_2 = \qsrt ( v_r + v ) / (v_r -v)$

Given data states that the man cross a river by starting at a point with \theta angle to the vertical and ends exactly opposite direction that means the horizontal velocity of the man while swimming corresponds to the ground be zero.

Rest the velocity of man with respect to river be v and velocity of river be v r. therefore,

                   $v_r -v sin \theta  = 0.\\=> v_r = v sin \theta\\ => cos \theta = \sqrt {(1 - ( v ^2 / v_r^ 2 )}.$

Then the time taken to reach the bank be

                   $t_1 = d / v cos \theta = d / \sqrtv^2 - u^2.$

On covering  the equal distance on downstream, the velocity corresponds to ground will now be $v + v_r$ and distance be same i.e. d,

So the time taken be $t_2 = d / v + u$.

Therefore $t_1 / t_2 = \qsrt ( v_r + v ) / (v_r -v)$