Question

A man is coming down an incline of angle 30'. When he walks with speed 2root3 he has to keep hisumbrella vertical to protect himself from rain. The actual speed of rain is 5 m/s. At what angle withvertical should he keep his umbrella when he is rest so that he does not get drenched?​

Answer 1

Answer:

when man is at rest then he must have to hold his umbrella at 37 degree with the vertical

Explanation:

velocity of man is along the incline plane

so we have

$\vec v = 2\sqrt3(cos30 \hat i - sin30\hat j)$

$\vec v = 3\hat i - \sqrt3\hat j$

speed of rain is given as 5 m/s

now we know that velocity of man with respect to rain must be along Y direction as we know that man holds his umbrella vertical when he is moving on inclined plane

so we know that

$v_{rm} = v_r - v_m$

$v_{rm} = 5(sin\theta\hat i - cos\theta\hat j) - (3\hat i - \sqrt3\hat j)$

so we have

$5 sin\theta - 3 = 0$

$\theta = sin^{-1}(\frac{3}{5})$

$\theta = 37 degree$

so when man is at rest then he must have to hold his umbrella at 37 degree with the vertical

#Learn

Topic : Relative motion

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