Question

a person moving eastward with velocity of 4.8km/hr rain appers to fall vertically downward with the velocity of 64km/hr. find the actual speed and direction of velocity

(physics question)

Answer 1

Answer: speed and direction of rain is 8 km/hr southwest

Explanation:

Velocity of person, Vp = 4.8 km/hr east

                                      = 4.8 in positive x axis

Velocity of rain to be appeared to the person, Vrp = 6.4 km/hr ( in negative y axis)

                                Vrp = Vr - Vp

                                  Vr = √( Vrp^2 + Vp^2)

                                      = √64

                                      = 8 km/hr southwest

                               

Answer 2

ANSWER OF EACH PART IS GIVEN BELOW.

Explanation:

Given, velocity of man, $\vec{v_m}=4.8\ \^{i}$.

Velocity of rain with respect of man, $\vec{v_{rm}}=6.4\ \^{j}$ ( It should be 6.4 not 64 to get a reliable answer).

Now, we know, $\vec{v_{rm}}=\vec{v_r}-\vec{v_m}$ ( Here $\vec{v_r}$ is the velocity of rain.)

Therefore, $\vec{v_{r}}=\vec{v_{rm}}+\vec{v_m}$$=6.4\ \^{j}+4.8\ \^{i}.$

Therefore, $v_r^2=v_{rm}^2+v_m^2\\\\v_r=\sqrt{v_{rm}^2+v_m^2}=\sqrt{6.4^2+4.8^2}=8\ km/h.$

Now, direction of velocity of rain from vertical,

$tan\theta=\dfrac{v_m}{ v_{rm}}=\dfrac{v_m}{ v_{rm}}=\dfrac{4.8}{6.4}=0.75 \ .$

Therefore, $\theta=tan^{-1}(0.75)=36.87^o.$

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