The water drops fall at regular intervals from a tap $$5\,m$$ above the ground. The third drop is leaving the tap at an instant when the first drop touches the ground. How far above the ground is the second drop at that instant? (Take $$g = 10\,m/{s^2}$$  )

Solution :
Let $$t$$ be the time interval of two drops. For third drop to fall
$$\eqalign{ & 5 = \frac{1}{2}g{\left( {2t} \right)^2}\,\,\left[ {{\text{As}}\,u = 0} \right] \cr & {\text{or,}}\,\frac{1}{2}g{t^2} = \frac{5}{4}\,......\left( {\text{i}} \right) \cr} $$
Let $$x$$ be the distance through which second drop falls for time $$t,$$ then
$$x = \frac{1}{2}g{t^2} = \frac{5}{4}\,m\,\,\left[ {{\text{from}}\,{\text{Eq}}{\text{.}}\left( {\text{i}} \right)} \right]$$
Thus, height of second drop from ground
$$ = 5 - \frac{5}{4} = \frac{{15}}{4} = 3.75\,m$$