A man is coming down an incline of angle 30∘. When he walks with speed √3m/s he has to keep his umbrella vertical w.r.t. ground to protect himself from rain. The actual speed of rain is 5 m/s. At what angle with vertical should he keep his umbrella when he is at rest so that he does not get drenched?
θ=37∘
→vman=2√3cos30∘^i−2√3sin30∘^j
→vm=3^i−√3^j
At this moment the rain appears to him as vertical.Let the magnitude of |→vrm|=v
⇒→vrm=v^j
⇒→vrm=→vr−→vm
⇒→vr=→vrm+→vm
Draw vector diagram
⇒|→vr|=5
Given now from it we know
→vr=5 sin θ ^i−5cos θ‘^j
→vr=→vrm+→vm
⇒5 sinθ^i=−v^j+3^i−3^j
⇒(5sinθ=3) -------------(1)
−5cosθ=−(v+3)
From equation (1) we know sinθ=35
⇒θ=37∘