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Cloud is showering 2 water drops from 1000 meters above the ground in 1 second difference.

Let be Gravitational acceleration = $10\frac{m}{s^2}$, and Air resistance is negligible.

What will be the Time difference between the drops when the first hits the ground, and what will be the distance between the drops when the first hits the ground ?

This is what I have done,

$x(t) = x(0) + v_{0}t + \frac{at^{2}}{2}$

$\rightarrow 0 = 1000 - 5t^{2} \rightarrow t = 10\sqrt{2}$ for the first drop

$\rightarrow 0 = 1000 - 5(t+1)^{2} \rightarrow t = 10\sqrt{2} -1$ for the second drop.

And then, The time diffrence is $1$ second.

Then, I can evaluate $x(10\sqrt{2} -1) = v_{0}t + \frac{at^{2}}{2}$, and find the distance between the drops is $136$ meters.

There is faster or better way to solve that ? Thanks.

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  • $\begingroup$ @DavidK Yes it is, Sorry. $\endgroup$ – Noam Mar 18 '16 at 13:30
  • $\begingroup$ You use the same symbol, $t$, for different things without explaining, which I found confusing. But to the nearest meter, $136$ meters is what I get too. $\endgroup$ – David K Mar 18 '16 at 13:34
  • $\begingroup$ So, What should I do ? $\endgroup$ – Noam Mar 18 '16 at 13:37
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The question shows you already found the correct numeric result. I would just suggest working on the presentation so that people reading your calculations do not misunderstand them. (Eventually you may also be solving problems that are complicated enough that even you will not be able to remember what you're doing unless you explain it carefully.)

I would assign a different number or name to the time of each event. It often helps to choose a convenient event and measure all times relative to that event. So if the event is when the first drop starts falling, the first drop starts falling at $0$ seconds and the second drop starts falling at $1$ second. When you are first setting up the problem you have not yet calculated when the first drop hits the ground, so give that a name: you might say it happens at $t_1$ seconds. And you can say the second drop hits the ground at $t_2$ seconds.

So the first drop falls for a total of $t_1$ seconds, which gives a vertical distance of $$ 0=1000−5t_1^2.$$ From this is follows that $t_1 = 10\sqrt2$; this is really the same thing you already did, just giving a unique name to this time so it's clear which time we're talking about.

At time $t_1$, when the first drop hits the ground, the second drop has only been falling for $t_1 - 1$ seconds. So its height above the ground at that instant is $$1000 - 5(t_1 - 1)^2 = 1000 - 5(10\sqrt2 - 1)^2 \approx 136.$$

And of course at $t_2$ seconds, when the second drop hits the ground, the drop has been falling for $t_2 - 1$ seconds, and so $$ 0=1000−5(t_2 - 1)^2.$$ From this we can figure out that $t_2 = 10\sqrt2 + 1$; personally, however, I would just observe that since the two drops travel the same distance with the same initial velocity and acceleration, they take the same elapsed time, so $t_2 - 1 = t_1 = 10\sqrt2$ and we can immediately solve this to find that $t_2 = 10\sqrt2 + 1 = t_1 + 1$ without dealing with a quadratic.


If we wanted a more long-winded solution I suppose we could say that if the second drop has been falling for $\Delta t_{21}$ seconds when the first drop hits the ground, then $t_1 = \Delta t_{21} + 1$ (because the first drop started one second earlier) and therefore $$ 0=1000−5(\Delta t_{21} + 1)^2.$$ That's the equation $0=1000−5(t + 1)^2$ that you wrote, but writing $\Delta t_{21}$ instead of $t$ is a reminder that this equation was true with regard to a very specific amount of elapsed time. we can solve that equation to find out that $\Delta t_{21} = 10\sqrt2 - 1$, and then find the height of the second drop above the ground at time $t_1$ by evaluating $1000 - 5(\Delta t_{21} - 1)^2,$ but I'm already having more trouble remembering which symbol means what than I did in the previous solution, so I think maybe this isn't the easiest way to go.

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You can do it faster, by the concept of relative velocity and acceleration.

When the second drop starts, the first drop would have moved -> $\frac {at^2}2=5m$ and would have a velocity of $at=10m/s$

Now, the relative velocity of the first drop w.r.t second one is $v_2-v_1=10m/s$ and the relative acceleration is $0$, since both have same acceleration.

A water droplet takes $t$ time to fall. If a water droplet starts at $1$ sec, it will fall at $t+1$. So, time difference will still remain $1$.

Now, you know the time taken for the first drop to fall- $10\sqrt{2}-1$
Use the equation $x=ut+\frac 12 at^2$ where $a=0$ and $u=10$ (relative).

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  • $\begingroup$ But, How relative velocity and acceleration can help me to solve this ? $\endgroup$ – Noam Mar 18 '16 at 14:00
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To answer the first question no calculation is needed: the two drops take the same time to hit the ground, so if they start one second apart, they'll arrive one second apart.

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  • $\begingroup$ Ok,But if I want to prove it by math , How should I do this ? $\endgroup$ – Noam Mar 18 '16 at 13:48
  • $\begingroup$ Logic IS math... $\endgroup$ – Aretino Mar 18 '16 at 13:54
  • $\begingroup$ By the way: you have the wrong sign in your third formula. It should be $0=1000-5(t-1)^2$, whence $t=10\sqrt2+1$: the second drop arrives one second AFTER the first one. $\endgroup$ – Aretino Mar 18 '16 at 13:58
  • $\begingroup$ Why minus ? Can you explain it ? $\endgroup$ – Noam Mar 18 '16 at 13:59
  • $\begingroup$ How many seconds has the second drop travelled, if the first drop has fallen for $t$ seconds? $\endgroup$ – Aretino Mar 18 '16 at 14:01

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