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Find the maximum height?

  • Thread starter Math10
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  • #1
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Homework Statement


A stone weighing 1/2 lb is thrown upward from an initial height of 5 ft with an initial speed of 32 ft/s. Air resistance is proportional to speed, with k=1/128 lb-s/ft. Find the maximum height attained by the stone.

Homework Equations


None.

The Attempt at a Solution


Here's my work:
mg+kv=m(dv/dt)
(1/2)(-32)-(1/128)v=(1/2)(dv/dt)
(dv/dt)+(1/64)v=-32
integrating factor method:
e^(t/64)
(e^(t/64))v=-2048e^(t/64)+C
v=-2048+Ce^(-t/64)
v(0)=32
C=2080
v=-2048+2080e^(-t/64)
v(t)=h'(t)
h'(t)=-2048+2080e^(-t/64)
integrate h'(t),
h(t)=-2048t-133120e^(-t/64)+C
h(0)=5
C=133125
h(t)=-2048t-133120e^(-t/64)+133125
v(t)=0 when t=0.992268
h(0.992268)=20.8353
But the answer in the book says 17.10 ft. What's wrong? Is my answer correct? If not, then please correct me.
 

Answers and Replies

  • #2
130
30
Is there an angle for the initial speed or is it all horizontal?
 
  • #3
301
0
I don't know.
 
  • #4
130
30
Nevermind it says thrown upward
 
  • #5
33,262
4,963

Homework Statement


A stone weighing 1/2 lb is thrown upward from an initial height of 5 ft with an initial speed of 32 ft/s. Air resistance is proportional to speed, with k=1/128 lb-s/ft. Find the maximum height attained by the stone.

Homework Equations


None.

The Attempt at a Solution


Here's my work:
mg+kv=m(dv/dt)
(1/2)(-32)-(1/128)v=(1/2)(dv/dt)
The mass of the stone is not 1/2 lb -- that is its weight.
In the MKS system, mass is in kilograms, and weight is in newtons, but in the English system, mass is in slugs, and weight is in lbs.
Math10 said:
(dv/dt)+(1/64)v=-32
integrating factor method:
e^(t/64)
(e^(t/64))v=-2048e^(t/64)+C
v=-2048+Ce^(-t/64)
v(0)=32
C=2080
v=-2048+2080e^(-t/64)
v(t)=h'(t)
h'(t)=-2048+2080e^(-t/64)
integrate h'(t),
h(t)=-2048t-133120e^(-t/64)+C
h(0)=5
C=133125
h(t)=-2048t-133120e^(-t/64)+133125
v(t)=0 when t=0.992268
h(0.992268)=20.8353
But the answer in the book says 17.10 ft. What's wrong? Is my answer correct? If not, then please correct me.
 
  • #6
33,262
4,963
Is there an angle for the initial speed or is it all horizontal?
I don't know.
Nevermind it says thrown upward
Which implies to me, "thrown vertically upward."
 
  • #7
301
0
So what do I do?
 
  • #8
33,262
4,963
So what do I do?
Use the right weight and mass. You are given that the weight of the stone is 1/2 lb. That's your mg. What's the mass of the stone? You have m on the right side of your diff. equation.
 
  • #9
301
0
So (1/2)-(1/128)v=m(dv/dt)?
 
  • #10
130
30
I'm not sure where the mistake is. Perhaps, try writing it as a second order DE and solving for the position instead, and then use the derivative of that to get your velocity? Theoretically, both ways should give the same answer. If it it gives you something different you know you made a mistake.

So ΣF = ma
mg + kx' = mx''

Mark44 said:
Use the right weight and mass. You are given that the weight of the stone is 1/2 lb. That's your mg. What's the mass of the stone? You have m on the right side of your diff. equation.
Edit: try thy that first^
 
  • #11
33,262
4,963
So (1/2)-(1/128)v=m(dv/dt)?
Isn't the 1/2 lb. weight acting downwards? IOW, shouldn't that first term on the left side be -1/2?

If you know the weight (in lbs.) of something, how do you get its mass? You need to put in a value for m on the right side.
 
  • #12
301
0
So since weight=mg=1/2, what's g?
 
  • #13
33,262
4,963
See http://en.wikipedia.org/wiki/Slug_(mass [Broken]).
 
Last edited by a moderator:
  • #14
301
0
Oh yeah, so g=9.8 m/s^2, right?
 
  • #15
33,262
4,963
Oh yeah, so g=9.8 m/s^2, right?
No. That's g in the MKS system. Your problem uses English units - feet, pounds, etc.
 
  • #16
33,262
4,963
I'm not sure where the mistake is. Perhaps, try writing it as a second order DE and solving for the position instead, and then use the derivative of that to get your velocity? Theoretically, both ways should give the same answer. If it it gives you something different you know you made a mistake.

So ΣF = ma
mg + kx' = mx''
I don't see that this is any improvement. x' = v and x'' = v'. Why change a first-order DE to a second-order DE? Besides, I don't believe that Math10 has studied second-order DEs yet.
 
  • #17
301
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So is g=32?
 
  • #18
Ray Vickson
Science Advisor
Homework Helper
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So is g=32?
Use Google, or look in a physics book. Alternatively, you can convert from MKS to English units, using published conversion factors.
 
  • #19
301
0
I got it!!! Thank you so much for the help, guys!!
 

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