# Direction of Motion Question!! A steady wind is blowing with a speed of 36km/hour. from clouds moving horizontally with the wind, heavy raindrops fall to the ground 200m below. find the time taken...

Direction of Motion Question!!

A steady wind is blowing with a speed of 36km/hour. from clouds moving horizontally with the wind, heavy raindrops fall to the ground 200m below.

find the time taken for a drop to reach the ground

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### 5 Answers

This is a weird question because it might cause conflict conceptually with students. However, the horizontal velocity does not affect the vertical velocity in either case.

The physics of falling objects, *neglecting* drag, suggests that the object will take the same amount of time to fall regardless of mass. The reality is that raindrops are subject to drag and reach a terminal velocity rather quickly (within 1 sec or so). That problem seems a bit advanced for high school, since the solution requires programming or lots of iterative steps.

Neglecting air resistance, the rain drop, or any object, will be pulled to the Earth in approximately constant acceleration due to gravity. The time derived from the following would be the minimum time it would take for an object to fall 200 m, regardless of horizontal velocity.

`\Deltay = 1/2g*t^2 + v_(yi)t`

Where `g = 9.8m/s^2` , `v_(yi) = 0 m/s` and `∆y=200 m`

Thus, the above equation becomes

`t=sqrt((2\Deltay)/g): tgt0`

`t=sqrt((2(200\text(m)))/(9.8\text(m/s^2)))\rArr t=6.39 \text(s)`

The wind speed of 10 m/s horizontally only affects where it will land (63.9 m horizontally from where it started)

Initial vertical speed is 0, while horizzontal speed is 36kms/h

36 kms per hour are 36,000 meters per hours. since an hour is 3,600 seconds it menas the cloud have an speed of 36,000 meters in 3,600 seconds , that is 10 m/s

Now its horizontal speed that make path longer than it would be zero.

Speed of the rain is composed by horizontal speed (10 m/sec) and vertical speed gt:

`s= sqrt((s_h)^2+ (s_v)^2)` `=sqrt(g^2t^2+10^2)=` `sqrt(96.236 t^2+100)`

On the other side, if horizontal speed was zero, the rain fall in a time `t_0=sqrt(2h/g)` being zero initial vertical speed.

so :`t_0=6.385 sec ` ``

In `t_0`sec the coluds has moved on `s_ht_0` meters, that is `63.85` meters

It show, for for the principle of superimposition of the effects, the real path is: `sqrt(200^2+63.85^2)=209.944` meters

So at last we get the equation:

`209.44=t sqrt(92.636 t^2+100)`

`43,682=t^2(92.635t^2+10)`

```92.635 t^4+10t^2-43,682=0`

`Delta= 16,185,982.28`

`t^2= (-10+-4,023.173)/185.27`

`t_1^2=-21.769` nota accepted for a negative square.

`t_2^2=21.661`

so that : `t=4,654` sec.

Let drop take t seconds to reach the ground.

clouds are moving horizontally ,so we have initial velocity of the drop

is zero.

motion undergravitational force,

u=0 ,g =9.8 m/sec

`h=ut+(1/2) g t^2`

`240=(1/2)xx9.8 t^2`

`t^2=240/4.9`

`t=6.998`

`t=7` sec. approx

Sorry ,Please ,typographical mistake please change answer instead 240 ,it should be 200.

`h=ut+(1/2)g t^2`

`200=(1/2)xx9.8xxt^2`

`40.816= t^2`

`t=6.39`

Let drop take t seconds to reach the ground.

clouds are moving horizontally ,so we have initial velocity of the drop

is zero.

motion undergravitational force,

u=0 ,g =9.8 m/sec

sec. approx

wouldn't h= 200, since that is the height?