Kinematics problem: Drops of water fall from the roof of a building 9m high at regular intervals of time

Question

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 same instant the fourth drop starts its fall. What are the distances of the second and third drops from the roof? Take $g=10 m/s^2$

a) $4m$ and $1m$

b) $4m$ and $2m$

c) $6m$ and $2m$

Answer: 4m and 1m

My attempt: Let the interval between the falling of two drops be "t" seconds.

Then 4th drop falls after $3t$ seconds of falling of 1st drop. Therefore, the time of flight of 1st drop is 3t seconds.

$ H=\frac{1}{2}$ $gt^2$

From this equation, $t=\sqrt{\frac{1}{5}}$ seconds.

Distance of 2nd drop from the roof = $ H=\frac{1}{2}$ $g$ $\frac{1}{5}$ =$1m$

Distance of 3rd drop from the roof = $ H=\frac{1}{2}$ $g$ $\frac{4}{5}$ = $4m$

Hence, I get (a) as the answer.

Doubt: In this solution, I have assumed that the time difference between 1st and 2nd drop is always $\sqrt{\frac{1}{5}}$ seconds, i.e at any instant, it will only take $\sqrt{\frac{1}{5}}$ seconds for the second drop to reach the first drop. (This same logic goes with other drops too) I can't think of any explanation as to why the time difference will always be constant?

Can anyone please explain that?

Answer 1

You are correct. The time difference between successive drops reaching the same position, is constant. This is because the drops are released at regular intervals, as stated in the question.

$H=\frac12 gt^2$ : distance fallen is proportional to $t^2$. The drips are like a ticking clock. At each 'tick' of the clock, when the next drop is released, the drops have fallen distances of $1^2=1m, 2^2=4m$ and $3^2=9m$, because each drop takes 3 'ticks' to reach the ground. The times between $0-1m, 1-4m$ and $4-9m$ are all the same. The drops are accelerating so they travel longer distances in the same time.

If we took photographs at intervals of one 'tick', the photos would always look the same. From one frame to the next, each drop has advanced to the next position. This is true for all positions of the drops, not only when they are at $0, 1, 4$ and $9m$ from the point of release.

Answer 2

I am not sure what you mean by the time difference between the first and second drop at any instant of time - this has no bearing on the problem. The distance each drop travels is quadratic in the elapsed time it has been falling (until it hits the ground, at which point the journey is over).

The time $t=\sqrt{\frac{1}{5}}$ is the time interval between the release of each drop, not just the first and second. It's given in the problem statement that this does not change (regular intervals).

Since the fourth drop is dropped at the instant the first drop reached the ground, the time that elapsed is equal to the time for three drops to have been released. If the first drop was released at time 0, the fourth was then released at time $3t$ as you correctly deduced.

So the time of flight of the first drop is then $T_f = 3t$

You also correctly worked out that $t=\sqrt{\frac{1}{5}}$. This can be deduced using the distance of travel of the first drop from roof to ground, $H = 9m$, to calculate the value of its time of flight in terms of $t$.

Given the approximation $ g = 10 m/s$, it is then easy to work out the distances traveled by the second and third drop from $S= {\frac{1}{2}gt^2}$. Their total travel times at the time the first drop reaches the ground are $2t$ and $t$ respectively.

The distance traveled by the second drop is then $S_2 = 5(2t)^2 $ or $20t^2$. The distance traveled by the third drop is $5t^2$. Plugging in the value of $t$ gives the final answer.

Answer 3

Ok. This is what came to my mind. And I hope this helps you.

Let us consider the system for your question. Its a defective tap. Defective in the way that even though its completely closed, some defect makes it to leak out water through its nosal at a rate $ v $ volumes per second ( $ v $ being very small as compared to that of the volume of a water drop ). This rate of water coming out would remain constant until and unless there is an increase in defect or there is an effective variation in the water supply to the tap. Your question too is set up through such a system.

Now, initially, this small volume of water starts accumulating at the opening of the nossal as the surface tension in there blocks their freedom to flow out.

This process of accumulation continues till the weight of the total volume of accumulated water is large enough to overcome the surface tension, hence leaving that total amount of accumulated water to leave out as a single drop. And let this volume of the water drop be $ V $ .

So, as the tap provides $ v $ volumes of water per second, let the time taken for the accumulation of $ V $ volume of water, which is the condition for the formation of each drop, be $ t $. So, each drop takes time $ t $ to form and this will be the time interval between successive drops and this will remain constant. After leaving the tap, the time taken by each drop to reach the ground would be the same as their downward acceleration would be the same ( $ g $ ). I guess this answers your question.