Mechanics> A man running on a horizontal road at 8km...

A man running on a horizontal road at 8km/h finds the rain falling vertically. He increases his speed to 12km/h and finds that the drops make angle 30 degree with the vertical. Find the speed and direction of rain with respect to the road.

Please help me out in making a diagram for this one.

Thanks.

Shivansh mathur , 9 Years ago

Grade 11

3 Answers

ADITI SINGH

Since when the person runs at 8 km/h the rain seems to be coming verticaly down thus the horizontal component of rain, Vx is 8 km/h

When he incrreases his relative speed by 4 km/h with the horizontal component of rain, it seems to fall at an angle of 30 degrees with the vertical.

It means the vertical component, Vy is: 4/Vy = tan 30

i.e.

Vy = 6.93 km/h

V = 10.58 km/h

and angle of falling is: tan^-1(8/6.93)

Last Activity: 9 Years ago

Shyam Ardeshna

When the man runs at 8 km/hr ☔️ rain falling along y axis and when he runs at 12 km/hr the angle with y axis is 30 degrees

So by relative velocity we can stop the man in case 1 and decrease the speed in case 2 by 8km/hr.

So now relative speed is 4 km/hr along horizontal and angle with vertical is 30 degrees.

Tan 30 = x/4

Calculate (x) which is speed of rain wrt ground

Answer is 4*root3

Last Activity: 8 Years ago

Rishi Sharma

Dear Student,
Please find below the solution to your problem.

This shows that the horizontal component of rain =8 km/hr
as the relative velocity of the rain has only a vertical component with respect the man.
Vx=8 km/hr
2nd case →VRM=VR−VM=(8↑−Vy↑)−12↑=−4↑−Vy↑
tan30o = 4/Vy
⇒ 1/root(3) = 4/Vy
Vy = 4root(3) km/hr
⇒∣V∣ =4root(3) km/hr
Velocity of Rain = 8 km/hr↑−4root(3) km/hr↑
⇒ tanα = 2/root(3) from vertical

Thanks and Regards

Last Activity: 5 Years ago