Easy-ish Conservation of Energy question

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Homework Help Overview

The problem involves a waterfall with a change in elevation of 440 meters, where the speed of water after falling 12% of the way is given as 93 m/s. The task is to determine the speed of the water at the top of the waterfall, using conservation of energy principles while neglecting air resistance and fluid friction.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy equations, questioning the correctness of their calculations and assumptions regarding the initial speed of the water.

Discussion Status

Some participants confirm the approach taken by the original poster, while others express concern about discrepancies between calculated results and textbook answers. There is an ongoing exploration of potential errors in the textbook's provided answer.

Contextual Notes

Participants note that the textbook answer appears to be inconsistent with the calculations, raising questions about the reliability of the textbook information.

elasticities
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Homework Statement


A waterfall has a change in elevation of 4.4*10^2 m. When the water has fallen 12% of its way to the bottom, its speed is 93m/s. Neglecting air resistance and fluid friction, determine the speed of the water at the top of the waterfall.

Homework Equations


Et1=Et2
Ek1+Eg1=Ek2+Eg2
1/2m(v1)^2+mgh1=1/2m(v2)^2+mgh2

The Attempt at a Solution


g=9.81 N/kg
v1=?
v2=93m/s
h1=440m
h2=440m*0.88=387.2m

1/2m(v1)^2+mgh1=1/2m(v2)^2+mgh2
1/2(v1)^2+gh1=1/2(v2)^2+gh2
1/2(v1)^2+(9.81)(440m)=1/2(93)^2+(9.81)(387.2m)

Am I doing it right so far?
 
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So far, so good.
 
1/2m(v1)^2+mgh1=1/2m(v2)^2+mgh2
1/2(v1)^2+gh1=1/2(v2)^2+gh2
1/2(v1)^2+(9.81)(440m)=1/2(93)^2+(9.81)(387.2m)
1/2(v1)^2+(4316.4)=4324.5+3798.432
1/2(v1)^2=3824.532
v1=87.5m/s

But this is not the answer in the answers section of the textbook.
 
The answer I get is very close to yours. (Just differs in the third digit.)

What textbook, by the way?
 
Doc Al said:
The answer I get is very close to yours. (Just differs in the third digit.)

What textbook, by the way?

Maybe the answer is wrong in the textbook, it says 5m/s.
It's Nelson Physics 12 from 2003 I think.
 
elasticities said:
Maybe the answer is wrong in the textbook, it says 5m/s.
It's Nelson Physics 12 from 2003 I think.
Sounds like the book messed up. If it started out at 5 m/s and fell the full 440 m, then it would have a speed of 93 m/s at the bottom.
 
Doc Al said:
Sounds like the book messed up. If it started out at 5 m/s and fell the full 440 m, then it would have a speed of 93 m/s at the bottom.

Silly textbook, thanks though! :)
 

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