Water drops fall at regular intervals from a tap which is 5 m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant

The correct option is 3.75 m.

Let the regular interval of time between the release of consecutive water drops be $t$.
At the instant the first drop touches the ground, the third drop is just leaving the tap. This means:
- The first drop has been falling for a duration of $2t$.
- The second drop has been falling for a duration of $t$.
- The third drop is just leaving the tap (time of fall = 0).

Let $g$ be the acceleration due to gravity. The distance fallen by the first drop in time $2t$ under free fall (starting from rest) is given by the second equation of motion:
$H=\frac{1}{2}g{\left(2t\right)}^2$
Given that the tap is at a height of 5 m above the ground ($H=5\text{m}$):
$5=\frac{1}{2}g\left(4t^2\right)$
$5=2g{t}^2$
$g{t}^2=2.5$

The distance fallen by the second drop in time $t$ is:
$d_2=\frac{1}{2}g{t}^2$

Substituting the value of $g{t}^2=2.5$ into the equation:
$d_2=\frac{1}{2}\left(2.5\right)=1.25\text{m}$

Therefore, the height of the second drop above the ground is the total height of the tap minus the distance it has fallen:
$h=H-d_2$
$h=5-1.25=3.75\text{m}$