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Hole Punched in the Side of Container

  • #1
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Homework Statement


A hole is punched in the side of an 20 cm tall container full of water. To maximize horizontal distance of the water, how far from the bottom of the container should the hole be punched?


Homework Equations


??
[tex]\Delta[/tex]y = v0t + .5at2


The Attempt at a Solution


This problem deals with fluid dynamics, but I'm not sure which equation might apply... therefore, I'm not sure how to begin. Help?
 

Answers and Replies

  • #2
174
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Hi,
These question deals with Bernoulli which I didn't study yet, but there is another way to solve it.
If you would know the initial speed of the water the comes out from the container,will you be able to solve the question? or it's not the thing that bothers you?
well to get the initial velocity,There is a "non written law" or maybe it's not , anyway:
the initial kinetic energy of the drop equals to the potential energy the drop has in the height the hole was punched.
you can get initial velocity here. now try to get the distance as a function of h, and tell me how it goes, I'll try to guide You from there.
Good Luck
 
  • #3
rl.bhat
Homework Helper
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Applying Bernouli's equation find the velocity of the water coming out of the hole.
Initially the velocity is horizontal and it remains constant.. The moves in a projectile motion. Its vertical velocity increases. Find the it takes to reach the floor. The range of the water R is given by the product of horizontal velocity X time it takes to reach the ground. To find the condition for maximum range find dR/dh and equate it to zero.
 
  • #4
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Applying Bernouli's equation find the velocity of the water coming out of the hole.
Initially the velocity is horizontal and it remains constant.. The moves in a projectile motion. Its vertical velocity increases. Find the it takes to reach the floor. The range of the water R is given by the product of horizontal velocity X time it takes to reach the ground. To find the condition for maximum range find dR/dh and equate it to zero.
I know that Bernoulli's equation is:

P1 + .5[tex]\rho[/tex]v12 + [tex]\rho[/tex]gy1 = P2 + .5[tex]\rho[/tex]v22 + [tex]\rho[/tex]gy2

However, I am given only the height of the container (20cm). Would it be correct to say that
P1 = P2 = P0 and that v2 = O.

Then v1 becomes [tex]\sqrt{2gh}[/tex].

But I still do not know how to determine how far from the bottom of the container the hole should be placed.

(By the way, the course I'm in does not rely on calculus knowledge, so is there a different way to approach this without having to take a derivative?)
 
  • #5
rl.bhat
Homework Helper
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The distance form hole to ground is H - h.
Using the formula (H-h) = vot + 0.5gt^2, find the time to reach the water to the ground. Here vo is zero. Then x = v1*t

In projectile motion, when the range is maximum, what is the relation between range and maximum height?
Here x = R/2 and maximum height is (H - h)
 
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  • #6
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Then v1 becomes [tex]\sqrt{2gh}[/tex].
Oh, wait... do I want to solve for v1? Or is it v2?

Sigh... I still do not know how to determine how far from the bottom of the container the hole should be placed.

Any help would be appreciated.
________________________________________

Edit: Hmmm... just got your post, rl.bhat. Will try your suggestion =)
Your help is appreciated.
 
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  • #7
103
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The distance form hole to ground is H - h.
Using the formula (H-h) = vot + 0.5gt^2, find the time to reach the water to the ground. Here vo is zero. Then x = v1*t

In projectile motion, when the range is maximum, what is the relation between range and maximum height?
Here x = R/2 and maximum height is (H - h)
So:

(H - h) = v0t +.5gt2

v0 = 0

H - h = .5(9.8)t2

h = v1t

H - v1t = .5(9.8)t2

and v1t = [tex]\sqrt{2gh}[/tex]t

Thus:

H - [tex]\sqrt{2gh}[/tex]t = .5(9.8)t2


Is this correct?

Now I have two variables: height and time.

The relationship between range and maximum height: typically, wouldn't half of the range be the location of the maximum height? But in this case, we're starting from height h, such that h [tex]\neq[/tex] 0. Edit: Oh, wait! Is it potential energy? PE = mgh?
 
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  • #8
rl.bhat
Homework Helper
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In the projectile motion, when the range is maximum angle of projection is 45 degree and maximum range is equal to 4* maximum height.
In the problem H - h = 0.5gt^2.
Hence t = [2*(H-h)/g]^1/2
Now x = 2(H-h) = v1*t
Substitute the values of t and v1 and simplify.
 
  • #9
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Ok... will try that right now.
 
  • #10
103
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In the projectile motion, when the range is maximum angle of projection is 45 degree and maximum range is equal to 4* maximum height.
In the problem H - h = 0.5gt^2.
Hence t = [2*(H-h)/g]^1/2
Now x = 2(H-h) = v1*t
Substitute the values of t and v1 and simplify.
If x = v1*t, then:

t = [tex]\sqrt{(v1*t)/g}[/tex] = [tex]\sqrt{\sqrt{2gh}t/g}[/tex]

Plugging t into the equation gives:

H - h = .5(9.8)((t[tex]\sqrt{2gh}[/tex])/g)

Correct?
 
  • #11
rl.bhat
Homework Helper
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2(H-h) = (2gh)^1/2*[2(H-h)/g]^1/2 = 2[h(H-h)]^1/2
[H-h] = [h(H-h)]^1/2
Now simplify.
 
  • #12
103
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2(H-h) = (2gh)^1/2*[2(H-h)/g]^1/2 = 2[h(H-h)]^1/2
[H-h] = [h(H-h)]^1/2
Now simplify.
2(H - h) = 2[tex]\sqrt{h(H - h)}[/tex]

Square both sides:

H2 - 2Hh + h2 = Hh - h2

H2 - 3Hh + 2h2 = 0

Is this ok?
 
  • #13
rl.bhat
Homework Helper
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OK. Now solve for h.
 
  • #14
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2(H - h) = 2[tex]\sqrt{h(H - h)}[/tex]

Square both sides:

H2 - 2Hh + h2 = Hh - h2

H2 - 3Hh + 2h2 = 0

Is this ok?
If it is ok... then:

(H - h)(H - 2h) = 0

You have the system of equations:

H + h = 20 --> 2H + 2h = 40

H - 2h = 0

3H = 40

H = 40/3 = 13.3

Plug that in to get h = 6.7cm?

So, 6.7 is the height from the bottom of the container?
 
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  • #15
rl.bhat
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In the problem H = 20 cm = height of the container. h is the height of the hole from the surface of the water.
 
  • #16
mukundpa
Homework Helper
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I think we have to maximize the horizontal distance covered by water with h to get min h
 
  • #17
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rl.bhat:
Correct me If I'm wrong , It's not a projectile motion, there is no angle, it is "shot" horizontally.
well anyway, My answer is not 6.7.
initial speed = LaTeX Code: \\sqrt{2gh}
now the initial height is : H-h, h is the distance from the top to where the hole is punched.
Using kinematics You should get D(distance)=Vi*Sqrt[2(H-h)/g)]
put Vi to the equation, and you get a function D by h.
Now How will You get the maximum distance of a function of distance??(hint:math!!!!! 2 ways to find the h,using derivative or by a law in math, that sqrt(m*n)<=(m+n)/2 always! , and sqrt(m*n)=(m+n)/2 when m=n, where do You see that in our problem?)
Good luck,Tell me how it goes,I hope I'm right anyway :P
 
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  • #18
rl.bhat
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Anything "shot" horizontally falls freely under gravity. What is the nature of this motion?
 
  • #19
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Anything "shot" horizontally falls freely under gravity. What is the nature of this motion?
Well I don't know how It's called, I'm not from the USA , here we divide the motion to a motion with an initial angle and a motion with initial angle =0.
Excuse me if that's what You meant,but You wrote in one of Your post that the maximum distance is in a 45 degrees angle, but where do You see an angle here?
 
  • #20
rl.bhat
Homework Helper
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Well I don't know how It's called, I'm not from the USA , here we divide the motion to a motion with an initial angle and a motion with initial angle =0.
Excuse me if that's what You meant,but You wrote in one of Your post that the maximum distance is in a 45 degrees angle, but where do You see an angle here?
At what initial angle of projection the range of an object is maximum? And is there any relation between this maximum range and maximum height?
In the above problem I am referring later half of the motion.
 
  • #21
174
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At what initial angle of projection the range of an object is maximum? And is there any relation between this maximum range and maximum height?
I don't think we understand each other, Of course the maximum range is when shot in 45 degrees,but i Don't think that It's related to the problem, The solution I wrote is IMO correct, this is what i think happens in the problem, maybe I understood it Wrong.
What do You think?
 

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