Distance Between Dropped Balls – Rankers Physics
Topic: Kinematics
Subtopic: Motion Under Gravity

Distance Between Dropped Balls


Assertion (A): Two balls are dropped one after the other from a tall tower. The distance between them increases linearly with time (elapsed after the second ball is dropped and before the first hits ground).
Reason (R): In given situation relative acceleration is zero, whereas relative velocity is non-zero.
 
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Solution:

Assertion (A): Let \(\Delta t\) be the time interval. The distance between them \(D(t) = \frac{1}{2}gt^2 - \frac{1}{2}g(t-\Delta t)^2 = g t \Delta t - \frac{1}{2}g (\Delta t)^2\). This is a linear function of time \(t\). So (A) is true.


Reason (R): Both balls accelerate at \(g\). Thus, their relative acceleration is \(\vec{g} - \vec{g} = \vec{0}\). The first ball has velocity \(g\Delta t\) when the second is dropped, so the relative velocity is non-zero and constant. So (R) is true.
(R) correctly explains (A): constant non-zero relative velocity results in linear increase in relative distance.

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