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TopNewestLet us derive a simplistic formula for terminal velocity:

Assume a spherical liquid drop of radius \(r\), density \(\rho\) and coefficient of viscosity \(\eta\). It is falling through air of density \(\sigma\).

The forces acting on it are:

At terminal velocity, our drop is in equilibrium:

\[\dfrac{4}{3} \pi r^3 \rho g = 6 \pi \eta r V_T + \dfrac{4}{3} \pi r^3 \sigma g\]

Simplifying this, we obtain the expression for terminal velocity:

\[\boxed{V_T = \dfrac{2 r^2 g}{9 \eta} (\rho -\sigma)}\]

This shows that a drop with larger radius will reach the ground at a higher terminal velocity. – Raj Magesh · 2 years, 1 month ago

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We can have a bonus question now..

If two identical drops fall from clouds at different heights, which will hit the groud with greater speed, the one from the higher cloud or the other one? – Rohit Gupta · 2 years, 1 month ago

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– Raj Magesh · 2 years, 1 month ago

Assuming they both manage to reach terminal velocity (a very reasonable assumption), they will both hit the ground at the same speed (terminal velocity).Log in to reply

– Rohit Gupta · 2 years, 1 month ago

Correct!! Bravo..!!!Log in to reply

Assuming drops to be spheres, just use that odd equation for terminal velocity of bodies. – Raghav Vaidyanathan · 2 years, 1 month ago

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– Rohit Gupta · 2 years, 1 month ago

If drag force is more then weight is more as well... Which one is gonna dominate and why??Log in to reply