Question

The water drops fall at regular intervals from a tap 5m above the ground ,the third is leaving the tap at instant the first drop touches the ground ,how far above the ground is the second drop at that instant

Answer 1

Answer:

Explanation:

Given:

Height =h=5m

To calculate time taken by first drop to reach ground:

u=0m/s, t=T

so,

s=ut+1/2at²

5=0+1/2aT²-----(1)

Time taken by second drop will be T/2

Therefore, its distance from tap

h=0+1/2(aT²)/4       ------------(2)

divide equ (ii) by equ (1)

h/5=1/4

h=1.25m

therefore H-h=5-1.25=3.75m

Answer 2

Answer:

3.75m`

Explanation:

`t=sqrt((2h)/g) =sqrt((2xx5)/10)=1s`

Let `t_0` is the interval between two drops. Then

`2 t_0=t`

`:. t_0=0.5s`

`2^(nd)` drop has taken `t_0` time to fall. Therefore distance

fallen,

`d=1/2 gt_0^2=(1/2) (10)(0.5)^2`

`=1.25 m`

`:.` Height from ground =`h-d`

`=5-1.25`

`=3.75m`