Question

A raindrop falling from a height ℎ above ground, attains a near terminal velocity when it has fallen through a height . Which of the diagrams shown in figure correctly $\left(\frac{3}{4}h\right)$ shows the change in kinetic and potential energy of the drop during its fall up to the ground?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Formula used:

$KE=\frac{1}{2}m{v}^2$

$PE=mgh$

At height h from ground raindrop has maximum potential energy. Kinetic velocity will be zero as it is dropped, its velocity will be zero. Then as the rain drop falls its 𝑃𝐸 starts decreasing and kinetic energy starts increasing.

The total mechanical energy will remain conserved if we neglect the air resistance. If there is some air resistance, there is some force called upthrust (in fluids) which opposes its motion. It depends upon velocity of object. As the velocity increases, upthrust also increases. Hence during fall of raindrop, first its velocity increases and then becomes constant after some time.

This constant velocity is called terminal velocity, hence 𝐾𝐸 will also become constant. 𝑃𝐸 decreases continuously as the drop is falling continuously. The variation in 𝑃𝐸 and 𝐾𝐸 is best represented by (𝑏).



Final Answer: (𝑏).