Time for object in free fall to reach bottom of reservoir

In summary, the conversation discusses a problem involving the motion of an object dropped from a height into a liquid filled reservoir. The conversation includes equations and attempts at solving the problem using algebra and numerical methods. Suggestions are given for finding the time taken by the object to hit the bottom of the reservoir and understanding the theory behind the solution.
  • #1
sciencegem
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I'm sorry I'm sure this question is basic, but hours later I'm still stumped. Any hints appreciated. Also I'm trying to work on thinking about the theory behind the solution, so if anyone can include the physics "why it works" behind their hint that would be greatly appreciated :)

Homework Statement



An object of unit mass is dropped from a height of 20 above a liquid filled reservoir of depth 50. If the acceleration of gravity is 9.l8 and the-resistance-to-motion coefficients of air and liquid are 1 and 4, respectively, compute the time taken by the object to hit the bottom of the reservoir.

Homework Equations



(I derived these from ma=mg-mρ , but I checked I was correct in the book I'm independently studying that the problem is from).

(1) y(t)=(mg/ρ)t + (m/ρ)[(m/ρ) - v0](exp{-(ρ/m)t} - 1) + y0

(2) v(t)=(mg/ρ) + [v0 - (mg/ρ)]exp{-(ρ/m)t}

ρ=resistance-to-motion coefficient


The Attempt at a Solution



My "attempt at a solution" has taken up four pages of my notepad so I'm not even going to bother trying to put it down here. What I'd like to do is solve for t1 using (1), plug that into (2) and use that as v0 when solving for t2 using (1). I run into two problems: first, (1) has the variable t as both a 'normal variable' and exponent (I'm sure my terminology is wrong but you get the point--I don't know how to solve for 't' when the variable is in two very different places, one being an exponent). I've tried to surmount that obstacle by writing exp{-(ρ/m)t} in terms of the other characters (again, sorry for my terminology), say using (2) and the information connected to t1 to get

exp{-(ρ/m)t} = -(v(t)-g)/g ,

and plugging that bag into (1), but then I end up with two unknown variables (in this case, t1 and v(t)), and I'm not sure what to do with them. I've tried solving for different things and plugging several different equations into each other, but I'm stumped. I'm sure I'm missing something very simple but sadly I very am a slow learner and, as before mentioned, need to work on basing my solutions on theory and logic rather than what I can plug into what. Anyways, thanks to anyone who read all that, and like I said hints appreciated.
 
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  • #2
I think you mean ma = mg - vρ.
Your method looks fine. You cannot solve the equation for t algebraically. You must use a numerical method.
Try starting with y(t)=(mg/ρ)t + y0 to produce a value for t, then plug that value into the exponential term of the full equations and iterate. It will either converge or diverge. If it diverges, try the converse approach, plugging iterated values into the (mg/ρ)t term.
 
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  • #3
sciencegem said:
I
(I derived these from ma=[STRIKE]mg-mρ[/STRIKE] ma=mg-vρ, but I checked I was correct in the book I'm independently studying that the problem is from).

(1) y(t)=(mg/ρ)t + (m/ρ)[(m/ρ) - v0](exp{-(ρ/m)t} - 1) + y0

(2) v(t)=(mg/ρ) + [v0 - (mg/ρ)]exp{-(ρ/m)t}

ρ=resistance-to-motion coefficient

You can transform the equation ma=mg-vρ to an equation for v(y) (y is the downward displacement)

ma=mdv/dt = m (dv/dy) (dy/dt )→ m vdv/dy =mg-ρv.

In air, v1(0)=0 and the final displacement is y1=20 m:

You can calculate the final v1 from the solution for v(y) with some iterative method, as haruspex suggested.

Then solve the de dv/dt = g-vρ/m, and knowing the final v, you get t1.
In water you should take the buoyant force into account in principle, but the volume of the falling object is not given, so you can ignore it.
For the motion in the water, the initial speed is v1(final) and the displacement is 60 m. Now you can use the y(t) equation, as you know the initial velocity. You will find that the initial speed is higher than the terminal speed and most of the distance is traveled by nearly constant velocity.



ehild
 
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  • #4
Thanks haruspex and ehild ! I will give those suggestions a shot. I did indeed mean mg-vp silly me, thanks for the correction. The thing that perplexes me it that the book is on DEs and there has been absolutely no coverage of iterative methods yet--in fact the section this problem is from was on models with linear equations. I am glad to know though I wasn't missing something completely obvious. Thanks again!
 
  • #5
I think one thing you could do that will help you understand "the theory behind the solution" is to keep in mind that physical quantities have units associated with them. "height of 20", "depth 50", "acceleration of gravity is 9.8", and "the-resistance-to-motion coefficients of air and liquid are 1 and 4" are all meaningless statements because you did not include the units. Always state the units.
 
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  • #6
dauto said:
I think one thing you could do that will help you understand "the theory behind the solution" is to keep in mind that physical quantities have units associated with them. "height of 20", "depth 50", "acceleration of gravity is 9.8", and "the-resistance-to-motion coefficients of air and liquid are 1 and 4" are all meaningless statements because you did not include the units. Always state the units.

Thanks dauto , that's great advice and I appreciate your mentioning that. Unfortunately the problem didn't give me units to work with, it just said things like "of 20 units".
 
  • #7
sciencegem said:
(1) y(t)=(mg/ρ)t + (m/ρ)[(m/ρ) - v0](exp{-(ρ/m)t} - 1) + y0


Check your equation. A "g" is missing.


y(t)=(mg/ρ)t + (m/ρ)[(mg/ρ) - v0](exp{-(ρ/m)t} - 1) + y0

Ignore my hint in Post #3. t1 is easy to find with iteration using eq. 1. Substitute the given data and simplify as much as possible.


ehild
 
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  • #8
ehild said:
Check your equation. A "g" is missing.


y(t)=(mg/ρ)t + (m/ρ)[(mg/ρ) - v0](exp{-(ρ/m)t} - 1) + y0

Thank you again ehild ! I see where I did that. I actually had the 'g' in my notes it's just another thing I messed up when transferring it to my post lol.

ehild said:
Ignore my hint in Post #3. t1 is easy to find with iteration using eq. 1. Substitute the given data and simplify as much as possible.


ehild

OK I'll give that a try. Thanks again to everyone for all the help!
 

Related to Time for object in free fall to reach bottom of reservoir

1. How is the time for an object in free fall to reach the bottom of a reservoir calculated?

The time for an object in free fall to reach the bottom of a reservoir is calculated using the equation t = √(2h/g), where t is the time in seconds, h is the height of the reservoir in meters, and g is the acceleration due to gravity (9.8 m/s²).

2. Does the mass of the object affect the time it takes to reach the bottom of the reservoir?

No, the mass of the object does not affect the time it takes to reach the bottom of the reservoir. This is because the acceleration due to gravity is independent of the mass of the object.

3. What is the acceleration due to gravity?

The acceleration due to gravity is a constant value of 9.8 m/s², which represents the rate at which objects accelerate towards the Earth's surface due to the force of gravity.

4. Can the time for an object in free fall to reach the bottom of a reservoir be affected by air resistance?

Yes, the presence of air resistance can affect the time it takes for an object in free fall to reach the bottom of a reservoir. However, this effect is usually very small and can be neglected in most cases.

5. What other factors may affect the time for an object in free fall to reach the bottom of a reservoir?

Some other factors that may affect the time for an object in free fall to reach the bottom of a reservoir include the shape and size of the object, the density of the fluid in the reservoir, and any external forces acting on the object.

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