A tap has water drops dripping from it at a steady rate of 3 drops per sec. What is the vertical separation (in metres) between two consecutive drops, at an instant when the lower drop has a velocity of 9m/s.
(take `g=10m//s^(2)`)

Answer

`2.44`

Explanation

Time taken by each drop to come out of tap`=1/3S`
At a given instant if drop A is in motion for
`t_(1)` s then drop B is in motion for `(t_(1)-1/3)s`
For drop `A,u=0,v=9m//s^(2),s=h_(A)`
`a=10m//s^(2),t=t_(1)`

`:.v=u+at`
`:.9=0+10xxt_(1)`
`:.R=71.42m`
`s=ut+1/2at^(2)`
`:.h_(a)=0+1/2xx10xx(0.9)^(2)`
`:.h_(a)=4.05m`
For drop `B,u=0,s=h_(B),a=10m//s^(2)`
`t_(2)=(t_(1)-1/3)=0.9-1/3=17/30s`
`s=ut+1/2at^(2)`
`:.h_(B)=0+1/2xx10xx(17/30)^(2)`
`:.h_(B)~~1.61m`
Vertical separation between drop A and `B=h_(A)=h_(B)=-4.05-1.61=2.44m`