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zhuklara [117]
3 years ago
13

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
Physics
1 answer:
Elina [12.6K]3 years ago
6 0

Answer:

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

Explanation:

Mass of the rain drop = m = 0.5\times 10^{-4} kg

Weight of the rain drop = W

Duration of time = t = 3 seconds

W=m\times g

Drag force on rain drop = D=0.2\times 10^{-5} v^2

W=0.5\times 10^{-4} kg\time 10 m/s^2=0.5\times 10^{-3} N

Motion of the rain drop:

F=m\times a

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

W-D=m\times a

0.5\times 10^{-3} N-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times a

0.5\times 10^{-3} kg m/s^2-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times \frac{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\times 3 s

a=4.06 m/s^2

s=ut+\frac{1}{2}at^2 (second equation of motion)

s=0\times 3s+\frac{1}{2}\times 4.06m/s^2\times (3 s)^2

s = 18.27 m

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

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