when a man moves down an inclined plane with const. speed 5m/swhich make an angle 37 with the horizontal, he finds rain falling vertically downwards. when he moves up the same inclined plane with same speed , he finds the rain making an tan^-1(7/8)with horizontal then speed of rain is?

Dear sneha ,

Please see the figure to find the axis assumed.

let the x- component of rain be x and that of y be y . Now , apply the concepts of relative velocity , subtract the x component of man's velocity from rain 's xcomponent and same for y component of rain. 

now , man's x component = 5 *cos37 = 4 m/s , man's y component  = 5 sin37 = - 3 m/s

so , velocity of rain relative to man  : x component  =  x -(4 ) = (x - 4)   m/s ; y component  = (y +3) m/s 

as the velocity seems vertical horizontal component must be zero , so x - 4 =0 ,  x = 4 m/s

second case , man's x component = - 5 *cos37 =  -4 m/s , man's y component  = 5 sin37 = 3 m/s

so , velocity of rain relative to man  : x component  =  x -( -4 ) = (x + 4)   m/s ; y component  = (y -3) m/s

as angle from the horizontal is tan ^-1 (7/8)  , ; 7/8  = -( y- 3) / x +4 

7/8 = 3-y /8 , therefore , y = -4 m/s , i.e 4m/s downwards .

so velocity of rain is (4² +4² )^1/2  =  5.65 m/s   45deg to the vertical.

GOOD LUCK !!!

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I have another doubt:This may be a little simple for you... But please help me.. Not able to understand at all... And please give an appropriate hint for any future reference to such or similar type of questions:QUESTION:Rain is falling with a speed of 20 ms–1. A man rides a bicycle holds an umbrella making an angle of 300 with vertical. When he halves his speed, he has to hold the umbrella vertically for maximum protection. The initial speed of bicycle is   (a) 3 5 ms–1 (b) 10 ms–1 (c)  3 10 ms–1 (d) 20 ms–1