Drops of water fall at regular intervals from the roof of a building of height `h=16m`. The first drop striking the ground at the same moment as the fifth drop is ready to leave from the roof. Find the distance betweent he successive drops.

Explanation

Step I: Time taken by the first drop to touch the ground `=t=sqrt((2h)/(g))`
for `h=16m,t=sqrt((16(2))/(g))=4sqrt((2)/(g))`
Time interval between two successive drops is
`Deltat=((1)/(n-1))t=((1)/(4))t=sqrt((2)/(g))`
Where `n=` number of drops
Step II:
Distance travelled by `1^(st)` drop
`S_(1)=(1)/(2)g(4Deltat)^(2)=(1)/(2)g(16)((2)/(g))=16m`
Distance travelled by `2^(nd)` drop
`S_(2)=(1)/(2)g(3Deltat)^(2)=(1)/(2)g(9)((2)/(g))=9g`
Distance travelled by `3^(rd)` drop
`S_(3)=(1)/(2)g(2Deltat)^(2)=(1)/(2)g(4)((2)/(g))=4m`
Distance travelled by `4^(th)` drop
`S_(4)=(1)/(2)g(Deltat)^(2)=(1)/(2)g((2)/(g))=1m`
Distance between `1^(st)` and `2^(nd)` drops`=S_(1)-S_(2)=16-9=7`m
Distance between `2^(nd)` and `3^(rd)` drops`=S_(2)-S_(3)=9-4=5m`
Distance between `3^(rd)` and `4^(th)` drops `=S_(3)-S_(4)=4-1=3m`
Distance between `4^(th)` and `5^(th)` drops`=S_(4)-S_(5)=1-0=1m`