Question

The water drop falls at regular intervals from a tap $5m$ 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?

• A. 1.25 m
• B. 2.50 m
• C. 3.75 m
• D. 5.00 m

Answer 1

Answer: (C)

Height of tap $=5m$.
For the first drop,
$5=ut+\frac{1}{2}g{t}^2=\frac{1}{2}\times 10t^2=5t^2$
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 $0.5sec$
$=\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times (0.5{)}^2=1.25m.$
Therefore distance of the second drop above the ground $=5-1.25=3.75m$.