Question

Rain appears to fall vertically downward to a man moving with a
velocity of 4 m/sec. when he doubles his speed then the rain appears to
strike him at an angle of 45' from horizontal. Then what is the actual
velocity of rain drops?​

Answer 1

Explanation:

Initial velocity of the man, me = 4 m/s

Final velocity of the man, m'e = 8 m/s

m'm = m'e - me = 8 - 4 = 4 m/s

tan 45 = rm/m'm

rm = (m'm) tan 45 = 1

re = √(rm)^2 + (me)^2  = 4 √2 m/s

θ = tan^-1 (me / rm) = tan^-1 (1)

θ = 45° degree

Hence the speed of the raindrop is 4 √2 m/s and the angle is 45°.

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