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Shane Macguire Grade: upto college level
`        Consider the situation of the previous problem. The man has to reach the other shore at the point directly opposite to his starting point. If he reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance that he has to walk. `
3 years ago

Deepak Patra
474 Points
```										Sol. Velocity of man  V ⃗m = 3 km/hr
BD horizontal distance for resultant velocity R.
X-component of resultant Rx = 5 + 3 cos θ
t = 0.5 / 3sinθ
which is same for horizontal component of velocity.
H = BD = (5 + 3 cos θ) (0.5 / 3 sin θ) = (5+3 cos⁡θ)/(6 sin⁡θ )
For H to be min (dH/dθ) = 0
⇒ d/(d θ) ((5+3 cos⁡θ)/(6 sin⁡θ )) = 0
⇒ –18 (sin2 θ + cos2 θ) – 30 cos θ = 0
⇒ –30 cos θ = 18 θ cos θ = –18 / 30 = –3/5
Sin θ = √(1-cos^2⁡〖θ 〗 ) = 4/5
∴ H = (5+3 cos⁡θ)/(6 sin⁡θ ) = (5+3 (-3/5))/(6 x  (4/5)) = 2/3 km

```
3 years ago
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