Rain appears to fall vertically downward to a man moving with a velocity of 4m/sec. When he doubles his speed then the rain appears to strike him at an angle of 45 degree with the horiozontal. Then what is the actual velocity of rain drop?
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The speed of the raindrop is 4 √2 m/s and the angle is 45°.
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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Rain is falling vertically downwards with a velocity of 3 km/hr. a man walks in the rain with a velocity of 4 km/hr. the raindrops will fall on the man with a velocity of ?
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