Water drops fall at regular intervals from a tap which is 5m 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.

Given that

Water drops fall at regular intervals such that h = 5m

Time interval between first and third drop $\text{t}=\sqrt{\frac{\text{2h}}{\text{g}}}$

                Or

$\text{s}=\text{ut}+\frac{1}{2}{\text{at}}^{\text{2}}$

$\text{H}=\text{0}+\frac{1}{2}{\text{at}}^{\text{2}}$

$5=\frac{1}{2}{\text{at}}^{\text{2}}$

$5=\frac{1}{2}\times {\text{gt}}^{\text{2}}$

$5=\frac{1}{2}\times \text{g}{\left(\text{2t}\right)}^{\text{2}}\Rightarrow 5=\frac{1}{2}\times \text{g}{\left(\text{4t}\right)}^{\text{2}}$

$5=2{\text{gt}}^{\text{2}}\cdots \cdots \left(1\right)$

$5-\text{h}=\frac{1}{2}{\text{gt}}^{\text{2}}\cdots \cdots \left(2\right)$

$\frac{\left(2\right)}{\left(1\right)}\Rightarrow \frac{5-\text{h}}{5}=\frac{\frac{1}{2}{\text{gt}}^{\text{2}}}{2{\text{gt}}^{\text{2}}}$

$\frac{5-\text{h}}{5}=\frac{1}{4}$

$20-4\text{h}=5$

$4\text{h}=15\Rightarrow \text{h}=\frac{15}{4}=3.75\text{m}$