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Q.
The water drop falls at regular intervals from a tap $5 \,m$ above the ground. The third drop is leaving the tap at instant the first drop touches the ground. How far above the ground is the second drop at that instant?
Height of tap $=5 m$.
For the first drop,
$5=u t+\frac{1}{2} g t^{2}=\frac{1}{2} \times 10 t^{2}=5 t^{2}$
or $t^{2}=1$
or $t=1 \sec$.
It means that the third drop leaves after one second of the first drop, or each drop leaves after every $0.5 \sec$.
Distance covered by the second drop in $0.5 sec$
$=\frac{1}{2} g t^{2}=\frac{1}{2} \times 10 \times(0.5)^{2}=1.25\, m .$
Therefore distance of the second drop above the ground $=5-1.25=3.75\, m$.