The water drops fall at regular intervals from a tap \[\text{5 m}\] above the ground. The third drop is leaving the tap at the instant the first touches the ground. How far above the ground is the second drop at that instant?

A. \[\text{1}\text{.25 m}\]
B. \[\text{2}\text{.50 m}\]
C. \[\text{3}\text{.75 m}\]
D. \[\text{4}\text{.00 m}\].

Hint:In the given question as the water drops fall at regular intervals from a tap above the ground so the distance travelled by first drop will be the height of tap which is here. Initially the water drop has the velocity zero.

Complete answer:
Given,

Height of tap (s) \[\text{= 5 m}\]

Third drop is leaving the tap when the first drop touches the ground.

Time taken by third drop to reach at the ground \[\text{= t s}\]
And, \[\text{g = 10 m/}{{\text{s}}^{\text{2}}}\]

For the first drop
$\Rightarrow s=ut+\dfrac{1}{2}g{{t}^{2}}$
$\Rightarrow 5=0\times t+\dfrac{1}{2}\times 10\times {{t}^{2}}$
$\Rightarrow {{t}^{2}}=1$
$\Rightarrow $ t=1 sec

It means that the third drop leaves after one second of the first drop. It means the second drop will be in between both of the first and third drop so we can say that each drop leaves after every \[\text{0}\text{.5 sec}\].

So the distance covered by second drop in \[\text{0}\text{.5 sec}\]
$\Rightarrow s=ut+\dfrac{1}{2}g{{t}^{2}}$
$\Rightarrow s=0\times t+\dfrac{1}{2}\times 10\times {{(0.5)}^{2}}$
$\Rightarrow s=1.25\,m$

Therefore, at this instant the distance of second drop of water above the ground = total distance from tap of water to ground – distance covered in \[\text{0}\text{.5 sec}\] by second drop\[=\text{5}-\text{1}\text{.25}=\text{3}\text{.75 m}\]

Therefore, the correct choice is: (C) $\text{3}\text{.75 }\!\!~\!\!\text{ m}$

Note:
The tap of water height which is \[\text{5 m}\] should be considered as the distance travelled by the first drop of water. By which the time taken in travelling \[\text{5 m}\] distance by a water drop is calculated. Also need to calculate the distance covered by the second drop of water.