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Normal Force

A water drop ...

Question

A water drop of radius $1.5 m m$ is falling from height $1 k m$ having drag constant $0.5$ density of water drop is $1000 k g / m^{3}$ and density of air is $1.29 k g / m^{3}$ . Find the terminal velocity.

8.7 m / s

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7.8 m / s

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5.6 m / s

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4.3 m / s

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Solution

The correct option is B $7.8 m / s$

As an object falls through the atmosphere, there is air resistance acting on it. Air resistance is dependent on the velocity of the falling object, so as it speeds up, the force acting on it gets larger and larger.
Eventually, the force from air resistance will equal the force due to gravity. At that point the forces, acting in opposite directions, will sum to zero meaning that the object will travel at a constant velocity: terminal velocity.
The terminal velocity varies based on drag coefficients and masses.

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1 k m

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0.04 mm

1.9 \times 10^{- 5} {Ns/m}^{2}

1.2 {kg/m}^{3}

1000 {kg/m}^{3}

g = 10 {m/s}^{2}

1.8 x 10^{- 5} N - s / m^{2}

m^{3}

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T a k e g = 10 m / s^{- 2} (F o r c e o f b u o y a n c y d u e t o a i r i s n e g l e c t e d)

0.04 mm

1.9 \times 10^{- 5} {Ns/m}^{2}

1.2 {kg/m}^{3}

1000 {kg/m}^{3}

g = 10 {m/s}^{2}

1.8 \times 10^{- 5} N s / m^{2}

1.2 k g / m^{3}

10^{3} k g / m^{3}

g = 10 m / s^{2}

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