For a man running horizontally along east at spped of 4 km/hr, rain appears to fall vertically. If he doubles his speed the rain appears to hit him at an angle at 30o with vertical. The original speed and direction of rain will bea)4 km/hr, 30o with vertical towards eastb)4 km/hr, 30o with vertical towards westc)8 km/hr, 30o with vertical towards eastd)8 km/hr, 30o with vertical towards westCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

Given information
- A man is running horizontally along east at a speed of 4 km/hr.
- Rain appears to fall vertically for the man at this speed.
- When the man doubles his speed, the rain appears to hit him at an angle of 30o with vertical.

To find
- The original speed and direction of rain.

Solution
Let's assume the speed and direction of rain to be V and θ, respectively.

When the man is running at a speed of 4 km/hr,
- His velocity = 4 km/hr towards east
- The rain's velocity = V km/hr vertically downwards

As the rain appears to fall vertically for the man, the relative velocity of rain with respect to the man should be zero.
Therefore, the horizontal component of rain's velocity should be equal and opposite to the man's velocity.
- Rain's horizontal velocity = 4 km/hr towards west

Now, let's consider the case when the man doubles his speed, i.e., he runs at a speed of 8 km/hr towards east.
- His velocity = 8 km/hr towards east
- The rain's velocity = V km/hr downwards at an angle of 30o with the vertical

As the rain appears to hit the man at an angle of 30o with vertical, the vertical component of rain's velocity should be equal to the man's velocity component in the upward direction.
- Rain's vertical velocity = 8*sin(30) = 4 km/hr upwards

To find the horizontal component of rain's velocity, we can use the Pythagorean theorem.
- Rain's speed = sqrt((4)^2 + (V*cos(30))^2)
- Rain's horizontal velocity = sqrt((4)^2 + (V*cos(30))^2) towards west

As the relative velocity of rain with respect to the man should be zero, the horizontal component of rain's velocity should be equal and opposite to the man's velocity
- Rain's horizontal velocity = 8 km/hr towards east

Therefore, we can equate the two expressions for rain's horizontal velocity and solve for V.
- sqrt((4)^2 + (V*cos(30))^2) = 8
- (V*cos(30))^2 = 8^2 - 4^2
- V*cos(30) = sqrt(48)
- V = sqrt(48)/cos(30)
- V = 8 km/hr

Thus, the original speed of rain was 8 km/hr and its direction was towards east.

Answer: Option C (8 km/hr, 30o with vertical towards east)

Answer

Option C is correct.

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