Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8 m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.

Given, height of nozzle from floor, H = 9.8m Let, 

$t_1=$time taken by first drop to reach the ground and 

$t_2=$time taken $2^{2d}$ drop. $t_2=\frac{t_1}{2}$

                       
g = acceleration due to gravity $\left(9.8m{s}^{-2}\right)$

u = initial speed = 0

$\begin{array}{l}Since,Hut+\frac{1}{2}g{t}_1^2\therefore t_1=\sqrt{\frac{2h}{g}}\\ \Rightarrow t_1=\sqrt{\frac{2X98}{9.8}}=\sqrt{2}s\therefore t_2=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}s\end{array}$
Again from Eq. (i) Distance covered by $2^{nd}$ drop, $x=\frac{1}{2}g{t}_2^2$

$\therefore X=\frac{1}{2}X9.8X{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{1}{2}X9.8X\frac{1}{2}=\frac{9.8}{4}=2.45m$

and position of second drop from grounds,

$H_2=H_1–x=9.8–2.45=7.35m$