The water drops fall at regular intervals from a tap $5 \mathrm{~m}$ 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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ )

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Aicharya · Accepted Answer

Height of tap $=5 \mathrm{~m}$ and $(\mathrm{g})=10 \mathrm{~m} / \mathrm{sec}^{2}$. For the first drop,
$5=u t+\frac{1}{2} g t^{2}=(0 \times t)+\frac{1}{2} \times 10 t^{2}=5 t^{2}$ or $t^{2}=1$ or $t=1$
It means that the third drop leaves after one second of the first drop. Or, each drop leaves after every $0.5 \mathrm{sec}$. Distance covered by the second drop in $0.5 \mathrm{sec}$
$=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}=(0 \times 0.5)+\frac{1}{2} \times 10=(0.5)^{2}=1.25 \mathrm{~m}$
Therefore, distance of the second drop above the ground $=5-1.25=3.75 \mathrm{~m}$.

The water drops fall at regular intervals from a tap $5 \mathrm{~m}$ above the ground. The third drop is…

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