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Equation of velocity with quadratic drag, vertical throw

  • #1

Homework Statement



I am having trouble deriving the velocity equation for an objekt moving vertically upwards. the net force will be
[tex]m*dv/dt=-mg-kv^2[/tex]
where k is the drag. At the time t=0 the object's velocity wil be
[tex]v(0)=v_0[/tex]
 

Answers and Replies

  • #3
im not sure how they get to that equation, and why they are using y, which seems to be an height.
I have tried seperating the variables, and have come to this integral
[tex]-√(m/gk)∫1/(1+u^2 )du=∫dt[/tex]
where u is a substitution defined by
[tex]u=√(k/mg)⋅v[/tex]
which leads to this
[tex]-√(m/gk)⋅arctan(√(k/mg)⋅v)+c=t[/tex]
however I'm not sure if this is corrrect, and the constant c must be solved for the condition t=0 and v(0)=v_0. Furthermore this will have to be solved for v, because i need the velocity equation.
 
  • #4
haruspex
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im not sure how they get to that equation, and why they are using y, which seems to be an height.
I have tried seperating the variables, and have come to this integral
[tex]-√(m/gk)∫1/(1+u^2 )du=∫dt[/tex]
where u is a substitution defined by
[tex]u=√(k/mg)⋅v[/tex]
which leads to this
[tex]-√(m/gk)⋅arctan(√(k/mg)⋅v)+c=t[/tex]
however I'm not sure if this is correct, and the constant c must be solved for the condition t=0 and v(0)=v_0. Furthermore this will have to be solved for v, because i need the velocity equation.
That all looks correct. Can you not rearrange it easily enough into the form v = f(t)?
 
  • #5
for the start condition (t,v)=(0,v_0) i can only cook c down to this
[tex]c=√(m/gk)*arctan(√(k/mg)*v_0 )[/tex]
which is going to give a pretty nasty equation when inserted into the previous equation, that i want to solve for v?
 
  • #6
which i eventually solve to this for v
[tex]tan((t-√(m/gk)⋅arctan(√(k/mg)⋅v_0 ))/√(m/gk))/√(k/mg)[/tex]

or in a more pretty picture: http://imgur.com/V53iV

but I'm not sure if it's correct, or if it can be shortened
 
  • #7
haruspex
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arctan(v) = t+c
v = tan(t+c) = (tan(t)+tan(c))/(1- tan(t)tan(c))
v0 = tan(c)
v = (tan(t)+v0)/(1- tan(t)v0)
 
  • #8
I don't really follow you there, would you mind elaborating that?
 
  • #9
haruspex
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[tex]-√(m/gk)⋅arctan(√(k/mg)⋅v)+c=t[/tex]
[tex]arctan(√(k/mg)⋅v)=\frac{c-t}{√(m/gk)}[/tex]
[tex]√(k/mg)⋅v=tan(\frac{c-t}{√(m/gk)})[/tex]
[tex]=tan(\frac{c}{√(m/gk)}-\frac{t}{√(m/gk)})[/tex]
[tex]=\frac{tan(\frac{c}{√(m/gk)})-tan(\frac{t}{√(m/gk)})}{1+tan(\frac{c}{√(m/gk)})tan(\frac{t}{√(m/gk)})}[/tex]
At t=0:
[tex]√(k/mg)⋅v_0=tan(\frac{c}{√(m/gk)})[/tex]
[tex]√(k/mg)⋅v=\frac{√(k/mg)⋅v_0-tan(\frac{t}{√(m/gk)})}{1+√(k/mg)⋅v_0 tan(\frac{t}{√(m/gk)})}[/tex]
[tex]v=\frac{v_0-\frac{1}{√(m/gk)}tan(\frac{t}{√(m/gk)})}{1+√(k/mg)⋅v_0 tan(\frac{t}{√(m/gk)})}[/tex]
 

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