Question

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Answer 1

QUESTION:

A man walking due east at the rate of 2kmph. The rain appears to him to come down vertically at the rate of 2 kmph. The actual velocity and the directions of rainfall with vertical respectively are??

ANSWER:

v_Rv

R

= 2√2 kmph along South-east(45° with east).

ASSUMPTIONS:

v_mv

m

= Velocity of man.

v_rv

r

= velocity of rain with respect to man.

v_Rv

R

= original velocity of rain.

GIVEN:

v_mv

m

= 2 kmph

v_rv

r

= 2 kmph

TO FIND:

v_Rv

R

= ??

EXPLANATION:

v_Rv

R

will be the vector sum of v_mv

m

and v_rv

r

.

$$\begin{lgathered}v_R = \sqrt{ {v_r}^{2} +{v_m}^{2} } \\ v_R = \sqrt{ {2}^{2} +{2}^{2} } \\ v_R = \sqrt{ 8} \\ v_R = 2 \sqrt{ 2} \:kmph\end{lgathered}$$

THE ORIGINAL VELOCITY WILL MAKE 45° WITH BOTH THE VELOCITY OF MAN AND VELOCITY OF RAIN WITH RESPECT TO MAN.

Hence option A is correct.

Answer 2

Answer:

First convert into vector form

Taking direction of east as i

Velocity of Man = 2 i

Velocity of rain = 2 J

We know that

Vrm=Vr -Vm

2j = Vr -2 i

Vr = 2i + 2j

So the required actual velocity

= √ 2² +2²

= 2 √2 km/h

let θ is the direction of rainfall with vertical

tan θ = Vm/Vrm

tan θ = 2 /2=1

=> θ = 45°

So the required direction of rainfall with vertical is 45°