A raindrop falls from a height $h$ above the ground. Because of air resistance it reaches a near-terminal (constant) velocity after it has fallen a distance $(3/4)h$ — that is, when it is still a height $h/4$ above the ground. Consider how its kinetic energy (KE) and potential energy (PE) change with time during the fall until it reaches the ground. Which of the following graphs correctly represents these changes? (a) PE rises with time while KE falls with time. (b) PE falls from its initial value along a curve, while KE rises from zero and then levels off to a constant value once terminal velocity is reached (near height $h/4$). (c) PE and KE are straight lines that cross — PE falling linearly to zero and KE rising linearly — with no levelling off. (d) Both PE and KE follow curves whose sum is not constant, with KE rising throughout and PE dipping and rising.

> [!TLDR] > As the drop falls, PE decreases and KE increases. Once it reaches terminal velocity (after falling $3h/4$, i.e. at height $h/4$), its speed — and hence KE — stays constant while PE keeps falling to zero. Only the graph where KE levels off is correct: option **(B)**. ### Reasoning - **Before terminal velocity** (first $3h/4$ of the fall): the drop accelerates, so KE increases and PE decreases. - **After terminal velocity** (last $h/4$): the drop moves at constant speed, so $KE=\tfrac12 mv_{term}^2$ is **constant**, while it keeps losing height and PE continues to fall to zero at the ground. Energy is **not** conserved here because air drag dissipates energy — so KE does not simply mirror PE. ### Why the others are wrong …

A raindrop falls from a height h above the ground. Because of air resistance it reaches a near-terminal (const

NCERT Exemplar · Q15

h

(a) PE rises with time while KE falls with time.

(b) PE falls from its initial value along a curve, while KE rises from zero and then levels off to a constant value once terminal velocity is reached (near height

h/4

(c) PE and KE are straight lines that cross — PE falling linearly to zero and KE rising linearly — with no levelling off.

(d) Both PE and KE follow curves whose sum is not constant, with KE rising throughout and PE dipping and rising.

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Concept understanding — Terminal Velocity

The Intuition: Why a Falling Object Stops Accelerating

Drop a steel ball bearing into a jar of honey. You expect it to fall, but it doesn't crash down like it would in air. Instead, it sinks slowly and steadily, at a constant speed, almost from the moment you let go. That constant speed is the terminal velocity.

Why does it happen? Two forces are at work. Gravity pulls the ball down. But the honey resists its motion — that's viscous drag. At the start, the ball is moving slowly, so the drag is small. Gravity wins, and the ball accelerates. But as speed increases, drag grows. Eventually, drag becomes exactly as strong as the net downward pull (weight minus buoyancy). At that point, the net force is zero. No net force means no acceleration — the ball continues at whatever speed it has reached. That speed is terminal velocity.

Note

Terminal velocity is not a limit you approach from above. You reach it from below, as acceleration dies out.

The Precise Statement

For a small, smooth sphere falling through a viscous fluid at low speeds (so the flow is laminar, not turbulent), the drag force is given by Stokes' law:

F_{\text{drag}} = 6 \pi \eta r v

where $\eta$ is the fluid's viscosity, $r$ is the sphere's radius, and $v$ is its speed.

The other forces are:

Weight downward: $W = \frac{4}{3}\pi r^3 \rho g$ (where $\rho$ is the sphere's density)
Buoyant force upward: $F_b = \frac{4}{3}\pi r^3 \sigma g$ (where $\sigma$ is the fluid's density)

The net downward force (weight minus buoyancy) is:

F_{\text{net}} = \frac{4}{3}\pi r^3 (\rho - \sigma) g

At terminal velocity, drag equals this net downward force:

6 \pi \eta r v_t = \frac{4}{3}\pi r^3 (\rho - \sigma) g

Solve for $v_t$ :

v_t = \frac{2 r^2 (\rho - \sigma) g}{9 \eta}

v_t = \frac{2 r^2 (\rho - \sigma) g}{9 \eta}

What the Formula Tells You

Radius squared — a larger sphere falls much faster. Double the radius, quadruple the terminal speed.
Density difference — if the sphere is only slightly denser than the fluid, it falls slowly. If it's lighter than the fluid ( $\rho < \sigma$ ), $v_t$ becomes negative — it rises.
Viscosity — thicker fluids (higher $\eta$ ) give a lower terminal speed. Honey stops a ball far more than water does.

Watch out

This formula only holds for laminar flow (low Reynolds number). For a fast-falling sphere in a low-viscosity fluid like air, turbulence sets in and Stokes' law breaks down. Then terminal velocity follows a different law involving the square root of size, not the square.

A Concrete Example …

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As the drop falls, PE decreases and KE increases. Once it reaches terminal velocity (after falling $3h/4$ , i.e. at height $h/4$ ), its speed — and hence KE — stays constant while PE keeps falling to zero. Only the graph where KE levels off is correct: option (B).

Reasoning

Before terminal velocity (first $3h/4$ of the fall): the drop accelerates, so KE increases and PE decreases.
After terminal velocity (last $h/4$ ): the drop moves at constant speed, so $KE=\tfrac12 mv_{term}^2$ is constant, while it keeps losing height and PE continues to fall to zero at the ground.

Energy is not conserved here because air drag dissipates energy — so KE does not simply mirror PE.

Why the others are wrong …

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Why the others are wrong …

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