A man is going east with a velocity of 3 kmph. rain falls veryically downwards at a speed of 10kmph
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The velocity of the raindrops relative to the man has 2 components:
10km/hr downwards (vector AB)
3km/hr west (vector BC) (because he is going east and relative to his forwards direction the rain will appear to be moving backwards).
..A
.
CB (right angle at B)
The velocity of the rain relative to the man is the vector sum of these comonents (AC).
The angle the umbrella should point is direectly into the rain, along AC.
AC makes an angle to the vertical which is the angle at A
tanA = ||BC|| / ||AB|| = 3/10 = 0.3
A = tan⁻¹0.3 = 16.7°
Since there are 60mins in 1°, 0.7° = 0.7x60 = 42min
So the angle is 16° 42mins (16° 42'),
10km/hr downwards (vector AB)
3km/hr west (vector BC) (because he is going east and relative to his forwards direction the rain will appear to be moving backwards).
..A
.
CB (right angle at B)
The velocity of the rain relative to the man is the vector sum of these comonents (AC).
The angle the umbrella should point is direectly into the rain, along AC.
AC makes an angle to the vertical which is the angle at A
tanA = ||BC|| / ||AB|| = 3/10 = 0.3
A = tan⁻¹0.3 = 16.7°
Since there are 60mins in 1°, 0.7° = 0.7x60 = 42min
So the angle is 16° 42mins (16° 42'),
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