Question

A water tap leaks so that water drops fall at regular intervals. Tap is fixed 5 m above the ground. First drop reaches the ground and at that very instant a third drop leaves the tap. At this instant, the second drop is at a height of
A. 3 m
B. 4.5 m
C. 3.75 m
D. 2.5 m

Answer 1

Hint: The droplets fall under constant acceleration due to gravity g from a height 5 m. Find the total time required to fall from a height of 5 m. Divide this total time by 2 to find out the position of the second droplet.
Formula used:
For any body falling under the influence of g, irrespective of its mass, the time distance covered by it in time t is:
$s = ut - \dfrac{1}{2}gt^2$ .

Complete answer:
First, we must find out the total time taken by the drop from a height of 5 m, to reach ground i.e., to travel a distance s = 5 m vertically with initial velocity u = 0. Second law of motion gives:
$5 = 0 - \dfrac{1}{2}(10)t^2$
After simplifying this we get t= 1s.

Now, we are given that the droplets fall at regular intervals. So, if we assume t = 0s time for droplet 3 and t = 1s time for droplet 1, then for droplet 2 we must have:
$t = \dfrac{1 - 0}{3 - 1} = \dfrac{1}{2}$s.

So, we have the time and we need to find the distance the drop covers from the tap in this time frame falling freely under gravity. Again, upon using second law of motion we get:
$s = 0 - \dfrac{1}{2}(10)\dfrac{1}{2^2}$
$s = -\dfrac{5}{4}$m.
Therefore, the drop covers a distance of 1.25 m from the tap vertically. But we are supposed to find the height of the second drop so h = (5 - 1.25) m = 3.75 m.

So, the correct answer is “Option C”.

Note:
It might appear so that the second drop must be midway between the two drop i.e., at a height of 2.5 meters, but never commit such mistake of choosing option (D) as the correct answer as; (1) Drops are separated uniformly or at equal intervals in time and not in vertical distance and (2) Drops also do not travel with constant velocity to the ground but they are accelerated due to gravity. Also, notice the negative sign of the s we obtained it appears there because distances measured up from the ground are positive and down from the tap will be negative.