Question

Rain water is falling vertically downwards with a velocity v.when the velocity of the wind is zero water is collected at a rate r.when the wind starts blowing horizontally at a speed u the rate of collection of water im the same vessel is

Answer 1

rate of collection of rain water is defined as

$Q$ = velocity of rain vertically down * area of crossection

now initial when there is no wind blowing rain is falling vertically

so we have

$R = v*A$

so above is the rate at which water is collected inside the vessel.

now when wind is blowing horizontally with speed u

the velocity of rain will become

$v_{net} = \sqrt{u^2 + v^2}$

now the velocity makes and angle with vertical which is given by

$\theta = tan^{-1}\frac{u}{v}$

now the vertical component of velocity of rain is

$v' = v_net cos\theta$

$v' = \sqrt{u^2 + v^2}* \frac{v}{\sqrt{v^2 + u^2}}[\tex]</p><p>[tex]v' = v$

so there is no change in the vertical component of velocity of rain

so the rate of water collected in the container will be same as initial

$rate = R$

so rate of water collection will be same as initial = r

Answer 2

Answer:When the wind blows in the horizontal direction at constant speed, the water will gain a horizontal component of velocity in addition to the vertical acceleration, and instead of falling vertically downward, the water will fall in a parabolic path. However, the rate at which the water falls will not get affected because the change in water concentration in bucket would still remain the same.

As stated above, the water in both the cases fall from the same height and with same vertical component of initial velocity, therefore the time taken by the water to reach the bucket would not change. It is important to note that the wind just changes the trajectory of the water but not the time of its flight. Therefore the rate at which the water fills the bucket does not change.

THEREFORE THE RATE WOULD REMAIN SAME