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Question

A person standing on a road has to hold his umbrella at 60o with the vertical to keep the rain away. He throws the umbrella and starts running at 20ms1. He find that rain drops are hitting his head vertically. Find the speed of the rain drops with respect to (a) the road and (b) the moving person.
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Solution

vrg=vm+vmgvrm=vrgvmg=vrg+(vmg)
When man is not moving he is observing actual velocity of the rain.
vrain=vsin60o^ivcos60o^j
=32v^iv2^j(ms1) ...(i)
When the man starts running with speed 20ms1, rain appears to fall vertically as seen by man.
Hence, the velocity of rain with respect to man.
vr.m=vrainvman ...(ii)
Let the velocity of rain with respect to man has magnitude
|vr.m|=vv^j=(32v^iv2^j)20^i
v^j=(20+32v)^iv2^j ...(iii)
Comparing left side and right side terms,
20+32v=0v=403(ms1)
v=v2=12(403)=203(ms1)
Hence, the actual velocity of rain (from (i)),
vrain=32(403)^i12(403)^j(ms1)
=20^i203^j(ms1)
Hence, magnitude of actual velocity of rain,
|vr|=402(ms1)
And the magnitude of velocity of rain with respect to man,
|vr.m|=203(ms1)
In relative velocity, the observer observes the velocity of an object considering himself at rest.
Consider the example of a man sitting in a moving train and observes the objects outside situated on the ground.

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