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I am trying to solve my physics homework about a Rainwater tank. The tank has a diameter of $0,50m$ and a height of $1m$. At the bottom of the tank there's a hole with an area $A=1mm^2$. The tank is empty at t=0. It's a rainy day and the water increases by $0,2cm^3$ per second.

In a first step I showed that the speed of the water running out of the tank is equal to $v=\sqrt{2gh(t)}$ In a second step I should find out: "what is the maximum height the water can reach?"

Given that I tried by setting both velocities equal:

Speed of water going in: $v'= \frac{0,2cm^3}{s}$

Speed of water going out: $v=\sqrt{2gh(t)}$

Also I tried to give the equation of h(t), which should be $h(t)=\frac{v'*t}{V}$

And this is where I messed up , I guess:

I rearranged that to $v'=\frac{h*V}{t}$

Set $v'=v$ giving $ \frac{h*V}{t}=\sqrt{2gh} $

$\frac{h^2*V^2}{t^2}={2gh} <=> \frac{h*V^2}{t}={2g} <=> h= \frac{2gt}{V^2}$

But the problem is that I can not put in a value for t .

Other tries ended up in the same way.

Be aware that v and V might look the same in the above.

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  • $\begingroup$ What are you trying to find? $\endgroup$ – QuIcKmAtHs Jan 12 '18 at 12:50
  • $\begingroup$ the maximum height of the water $\endgroup$ – LurioTabasco Jan 12 '18 at 12:51
  • $\begingroup$ It would be useful to calculate the rate at which the water enters(by height). You have the volume, so you can divide it by pi r^2 to get rate of height increase. $\endgroup$ – QuIcKmAtHs Jan 12 '18 at 12:59
  • $\begingroup$ I have considered that at the beginning by $h(t)=\frac{v'*t}{V}$ but this is where my problem occurs since I bring the time into play $\endgroup$ – LurioTabasco Jan 12 '18 at 13:01
  • $\begingroup$ What is capital V? Why do you need to solve for t when you are only asked for height? You don't even use A (area). You use speed for both linear and volume. $\endgroup$ – paparazzo Jan 12 '18 at 14:09
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You are not asked to solve for $t$. You are asked to solve for $h$.

$v$ and $v'$ are not he same unit of measure. You mix linear with volume.

You ignore the area of the hole.

Rate of water going in: $r'= \frac{0,2cm^3}{s}$

Rate of water going out: $r=1mm^2*\sqrt{2gh}$

Set the two rates equal and solve for $h$

$\frac{0,2cm^3}{s} = 1mm^2*\sqrt{2gh}$

$\frac{0,2cm^3}{s*0,01cm^2} = \sqrt{2gh}$

$\frac{20cm}{s} = \sqrt{2gh}$

The units work out correctly

I will leave the rest to you
Hint square both sides

Solving for the $t$ in $h(t)$ would be a more complex

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  • $\begingroup$ After reviewing this, it becomes more and more obvious. I was totally on the wrong path at the beginning. Thank you. $\endgroup$ – LurioTabasco Jan 14 '18 at 19:31
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The question gave you the diameter of the tank for a reason. The radius would be 0.25m = 25cm. So you take $\frac{20cm^3}{\pi*25^2cm^2}$ and you get the height the water column increases each second. We observe that the rate of water flowing out increases over time, while the rate of water entering remains at a constant $\frac{20cm^3}{\pi*25^2cm^2}$. Plotting a graph, you find the area formed between the two lines, and their intersection of speeds at a certain time T.

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  • $\begingroup$ Just solve for it would be easier than plotting $\endgroup$ – paparazzo Jan 12 '18 at 14:15
  • $\begingroup$ The graph is for reference, it might be clearer to OP. $\endgroup$ – QuIcKmAtHs Jan 12 '18 at 14:16
  • $\begingroup$ Thanks for this one. Finally makes sense when using the area. $\endgroup$ – LurioTabasco Jan 13 '18 at 7:13

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