Question

A boy riding a bicycle with a speed of 4 m/s towards east direction, observes rain falling vertically downward. If he increases his speed to 8 m/s, rain appears to meet him at 45 to the vertical. The speed of rain with respect to ground is

Answer 1

speed of rain wrt ground = 4√2 m/s .. answer ..

let velocity of Rain wrt ground ...

Vr = Vx i - Vy j

initial velocity of boy...

V1 = 4 i

rel. velo. of rain wrt boy ..

Vr1 = (Vx - 4)i - Vy j

since, rain is falling vertically...

so, Vx - 4 = 0

=> Vx = 4 .............(1)

final velocity of the boy ...

V2 = 8 i

rel. velo. of rain wrt boy...

Vr2 = (Vx - 8)i - Vy j

given, angle = 45°

tan 45° = -Vy / (Vx - 8)

=> Vy = 4 .......................(2)

from (1) and (2) ...

Vr = 4 i - 4 j ......ans ..

|Vr| = 4√2 m/s...... ans...

Answer 2

Answer:4√2 m/s

Explanation:He is going 4m/s east and sees the rain falling vertically downwards which means that the horizontal component of the rain is equal to the speed of the boy in the same direction

So, x component is 4m/s in east direction

Now, he picks up speed so relative velocity of rain in the x direction is….

V rain - V boy = V rain w.r.t. boy

= 4 ( from above result ) - 8 = -4m/s

Or, 4m/s towards west

Now 45 ° ,this is only possible when the x component of rain is equal to the y component of rain i.e. vertical component of rain is 4 m/s and direction is downwards ( obviously )

So w.r.t. ground speed of rain is 4√2 m/s in the downward - east direction

Something like this

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