Water drops are falling in regular …

Ans. Height of tower = 9m

Initial Velocity of drop u = 0

Acceleration Due to gravity = -10m/s 

Let time taken by first drop to reach ground = t

Then, Using

{tex}S = ut +{1\over 2}at^2{/tex}

{tex}=>- 9 = 0\times t + {1\over 2}(-10)t^2{/tex}

{tex}=>- 9 = -5t^2{/tex}

{tex}=> t^2 = {9\over 5} => t = {3\over \sqrt 5}{/tex}

Now, the fourth drop is just leaving the tap, the 2nd and 3rd drop are somewhere midway, and the first drop is hitting the ground. And the time interval for this situation to occur is {tex}{3\over \sqrt 5}{/tex}sec.

Since it says the drops are falling at regular intervals, therefore, the time must be equally divided in that time interval of {tex}{3\over \sqrt 5}{/tex} sec, giving {tex}{1\over \sqrt 5}{/tex} second of intervals between the drops. 

So time taken by second  drop  is {tex} {2\over \sqrt 5} sec{/tex}   and time taken by third drop is {tex}{1\over \sqrt 5}sec{/tex}.

Distance covered by second drop = {tex}{1\over 2} at^2 = {1\over 2}\times 10\times {4\over 5}{/tex}= 4 m

Distance covered by third drop = {tex}{1\over 2} at^2 = {1\over 2}\times 10\times {1\over 5}{/tex}= 1m

So, 2nd drop and 3rd drop are 4m and 1m away from top of tower respectively.