Question

A raindrop of mass 0.5 * 10^-4 kg is falling verctically under the influence of gravity. The air drag on the raindrop is fdrag = 0.2 * 10^-5 v^2, where v is the speed of the raindrop. Find the displacement of the ain drop after 3 second, assume the raindrops starts from rest and use 10m/s^2 for the acceleration of gravity

Answer 1

The displacement of the air drop after 3 second is 18.27 m.

Step-by-step explanation:

Mass of the rain drop = m =
`0.5* 10^(-4) kg`

Weight of the rain drop = W

Duration of time = t = 3 seconds

`W=m* g`

Drag force on rain drop =
`D=0.2* 10^(-5) v^2`

`W=0.5* 10^(-4) kg\time 10 m/s^2=0.5* 10^(-3) N`

Motion of the rain drop:

`F=m* a`

Net force on the rain drop , F= W - D

`W-D=m* a`

`0.5* 10^(-3) N-0.2* 10^(-5) v^2=0.5* 10^(-4) kg* a`

`0.5* 10^(-3) kg m/s^2-0.2* 10^(-5) v^2=0.5* 10^(-4) kg* (v)/(t)`

`0.006v^2+0.05v-1.5=0`

v = 12.18 m/s

Initial velocity of the rain drop = u = 0 (since, it is starting from rest)

v=u+at (First equation of motion)

`12.18 m/s=0m/s+a* 3 s`

`a=4.06 m/s^2`

`s=ut+(1)/(2)at^2` (second equation of motion)

`s=0* 3s+(1)/(2)* 4.06m/s^2* (3 s)^2`

s = 18.27 m

The displacement of the air drop after 3 second is 18.27 m.