IIT JEE , AIEEE Forum - Physics , Mechanics

bagaria_24

when a man moves down a inclined plane with a constant speed 5m/s which makes an angle of 37 degree with the horizontal, he finds that the rain is falling vertically downward.When he moves up the inclined plane with same speed, he finds that the rain makes an angle

=tan^-1(7/8) with the horizontal. the speed of the rain is

a)sqrt(116)m/s

b)sqrt(32)m/s

c)5 m/s

d)sqrt(73)m/s

srini

Hi.. is the plane fixed?

srini

HI. is it 5m/s?

srini

Hi........plane fixed?

srini

?Hey. my post isn't apppearing?

srini

he........sorry

INDIAN_ARMY19890

what is horizontal here? the inclined planeon which man is running or the plane on which the in clined plane is situated?

INDIAN_ARMY19890

if horizontal plane means the plane where man is waiking i.e the inclined plane then we will get two equations
from first condition when man is waiking downward
we get 5+vcosx=1
vcosx=4.............(1)
from 2nd condition we will get
35+7vcosx=8vsinx...........(2)
solving 1 and 2 we will get v=5m\s(approx)

INDIAN_ARMY19890

plz reply wheather sol is right or wrong plz give some rating friends i have not get any rating yet.

krishna.gopal

Let Velocity of rain be Vx i + Vy j (i horizontal and j vertical)
i) In first part velocity of man is 4 i + 3 j
Relative velocity of rain wrt to man = (Vx-4) i + (Vy-3) j which is vertical

So Vx = 4 m/s

ii) In second part Relative velocity of rain wrt to man = (Vx+4) i + (Vy+3) j which is 8 i + (Vy+3) j
So ( Vy+3)/8 = -7/8
So Vy = -10m/s
So speed of rain = sqrt (116) m/s

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