Water drops from a leaky tap are falling, at regular time intervals, on the ground 125 cm below the tap. The first drop strikes the ground when the· sixth drop just begins to fall. What is the height from the ground· of the third drop when the first drop strikes the ground? Take g = 10 ms^-2 

The time t taken by the first drop to strike the ground is given by

    $h=\frac{1}{2}g{t}^2 \Rightarrow t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 1.25}{10}}=0.5 s$

 At this time, the sixth drop just emerges from the tap.

 Hence the time interval between consecutive drops = $\frac{0.5}{5}$= 0.1 s. Therefore, the time for which the third drop has fallen is

 $t_3=0.1+0.1+0.1=0.3 s$ The height through which the third drop falls in 0.3 s is

          $$\begin{gathered} h_3=\frac{1}{2}g{\left(t_3\right)}^2=\frac{1}{2}\times 10\times {\left(0.3\right)}^2 \\ =0.45 m=45 cm \end{gathered}$$

The height of this drop above the ground when the sixth drop just begins to fall = 125 - 45 = 80 cm.