Water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap, the instant the first drop touches the ground. How far above the ground is the second drop at that instant. `(g = 10 ms^-2)`

Options

A. 1.25m
B. 2.50m
C. 3.75m
D. 4.00m

Answer

C

Explanation

Time for downward jurnery
`t=sqrt((2h)/(g))=sqrt((2xx5)/(10))=1` sec
So for regular interval drops are falling at interval of `(1)/(2)` sec.
height of `2^(nd)` drop `=5-(1)/(2)xx10xx((1)/(2))^(2)`
`s_(1)-s_(2)=(g)/(2)(9-4)=4.9xx5=24.5m`