Question

A man walks at 5 km/h in the direction of a 20-km/h wind. Raindrops fall vertically at 6.2 km/h in still air.

Answer 1

Explanation:

Let us take the road along X-axis and the rain drops falling vertically along -Y axis i.e. there velocities are

Man as V(x) and Rain as V(-y) .

if one wishes to see the velocity of rain with respect to man . and as observer he does not consider his own velocity , so temporarily he must be brought to rest.

One must apply a equal and opposite velocity on the man. so, put a velocity vector V(-x) direction.

but the same velocity must be applied to the rain also otherwise the equivalence will be disturbed.

so to the observer (man) the rain will have two velocities 4 km/h along -Y axis vertically down and 3km/h along -X direction.

both are perpendicular to each other , therefore the resultant velocity of rain as it appears to man

V(rain) = Sqrt{ 4^2 + 3^2} = Sqrt{16 + 9 } = 5 km/h

and the angle theta at which it strikes the man from the vertical

will be

tan (theta) = 3/4 from the vector diagram of the two velocities and resultant.

Now we come to the second part- what is observed about the velocity of man with respect to observer on rain drops.

here again the rain drop will have to be made static observer by applying velocity in +Y direction of 4 km/h and equivalently a velocity of 4km/h on the man who is already moving in +X direction with vel. 3 km/h.

The two velocities are again perpendicular to each other, therefore the resultant of the two will be same

V(resultant) = Sqrt( 16 + 9) = 5 km/h

however the direction in which the rain drop observer will see this velocity will be making an angle theta with the horizontal X axis such that

Tan (theta)= 4/3 .