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Question

A man is coming down an inclined plane of angle 30 with speed 23 m/s. He has to keep his umbrella vertical to protect himself from rain. If the actual magnitude of velocity of rain is 5 m/s, at what angle with vertical should he keep his umbrella when he is at rest?

A
30
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B
45
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C
37
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D
53
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Solution

The correct option is C 37
While coming down the inclined plane man has to keep his umbrella vertical rain is falling vertically w.r.t man
let Velocity of rain w.r.t. man;Vr,m=x^j then

Vr,m=x^j=VrVm

Now velocity of man in vector form is
Vm=Vmcos30^iVmsin30^j =23cos30^i23sin30^j =3^i3^j

Now, from above
Vr,m=VrVmx^j=Vr(3^i3^j)Vr=3^i+(x3)^j ...(i)

Given: |Vr|=5 m/s
So.
5=32+(x3)216=(x3)2

x=(4+3) or x=(4+3)
We will take ve value of x as rain is falling vertically w.r.t man.
Vr,m=(4+3)^j ...(ii)
From equation (i) and (ii)
Vr=3^i+(x3)^j
Vr=3^i4^j

As rain is falling with angle 37 from vertical, man should hold umbrella at an angle 37 with vertical while at rest.

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