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:

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

ASSUMPTIONS:

$v_m$ = Velocity of man.

$v_r$ = velocity of rain with respect to man.

$v_R$ = original velocity of rain.

GIVEN:

$v_m$ = 2 kmph

$v_r$ = 2 kmph

TO FIND:

$v_R$ = ??

EXPLANATION:

$v_R$ will be the vector sum of $v_m$ and $v_r$.

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

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.

NOTE : REFER ATTACHMENT FOR DIAGRAM.

Answer 2

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