drops of water fall from the roof of a building 9m high at regular intervals of time, the first drop reaching the ground at the instant fourth drops starts toful.what are the distances of the second and the third drop from the roof.

hi Prateek,

Please find below the solution to the asked query:


As per the question the drops are falling at regular time interval.


height of the building is 9m

Time taken by the first drop to reach the groung is t sec

so its given that when the fourth drop is about to get detached from the building,

the first drop reaches the bottom covering a distance of 9 m.

so using the equation of motion:

S = ut + at²/2
with u = 0 m/s

substituting s = 9 and a = g = 10 m/s²

we have t = ((2 x 9)/10)^1/2 = 1.34   sec


Hence this time interval is divided by 3 as for regular time interval condition of falling of drops = 0.44 sec
so the time for which the 2nd drop is in air is = 1.34 - 0.44 = 0.9 sec
hence the distance travelled by it under gravity = gt²/2 = {10 x 0.9²)/2=  4.05 m
time for which the 3 rd drop is in air is = 1.34 -0.44- 0.44 = 0.46 sec

Hence the distance travelled by it under gravity is = gt²/2 = {10 x 0.46²)/2 = 1.058 m


so the difference in height from the roof of the building between 2nd and 3rd drop = 4.05- 1.058 = 2.99 meters


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