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A man starts running along a straight ro...

A man starts running along a straight road with uniform velocity observes that the rain is falling vertically downward. If he doubles his speed, he finds that the rain is coming at an angle `theta` to the vertical. The velocity of rain with respect to the ground is :

A

`uhati-utanthetahatj`

B

`uhati-ucotthetahatj`

C

`uhati+ucotthetahatj`

D

`(u)/(tantheta)hati-uhatj`

Text Solution

Verified by Experts

The correct Answer is:
C

`vec_(RM)=vecv_(R)-vecv_m`
`vecv_(RM)=vecv_(R)-uhati=(v_(RX)hati+V_(RY)hatj)-uhati`
Since `vecc_(RM) has only y component with respect to the man.
So, `V_(RX)-u=0`
`impliesV_(RX)=u`
After doubling the speed
`vecv_(Rm)=(uhati+v_(Ry)hatj)-2uhati`
`=-uhati+v_(Ry)hatj`
Given `tantheta=(u)/(v_(Ry))impliesv_(Ry)=ucottheta`
so `vecv_(R)=uhati+ucottheta(hatj)`
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Knowledge Check

  • A man running along a straight road with uniform velocity vecu=uhati feels that the rain is falling vertically down along - hatj . If he doubles his speed, he finds that the rain is coming at an angle theta with the vertical. The velocity of the rain with respect to the ground is

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    `2uhati+u cot theta hatj`
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    `ui + u sin theta hatj`

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    `(5)/(sqrt3)` km/h
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    `(5sqrt3)/(2)` km/h
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