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Quadratic Drag

  • Thread starter zeromaxxx
  • Start date
  • #1
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Homework Statement


A ball is thrown vertically upwards at speed v0. Assume drag force is proportional to v2.

a) Show that, while moving upwards, Newton's Second Law gives a = -g(1+v2/vt2) where vt is the terminal speed.

b) Take v0 = 3vt and solve for v(t), the time tmax at which it reaches max height and the maximum height it reaches. Express your results in terms of g and vt

Homework Equations


a = -g(1+v2/vt2)

v = -vt tanh (t/T), where T = vt/g

y = -vt2/g * ln [cosh (gt/vt)]


The Attempt at a Solution



I got part A of the question simple enough, I'm having trouble with part B. I got the expression for v(t) which was the integral of the first equation. I'm having trouble looking for tmax and the max height. It sounds simple enough but I can't grasp around the idea on how to do it.
 

Answers and Replies

  • #2
hotvette
Homework Helper
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Hint: what is v at the max height? Think calculus.
 
Last edited:
  • #3
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What is v at the max height? Think calculus.
v at max height should be 0 right?

so then 0 = -vt tanh (t/T)

It's how to deal with the equation that comes after.

tanh x = (ex - e-x)/(ex + e-x)

tanh (t/T) = (et/T - e-t/T)/(et/T + e-t/T)

0 = -vt[ (et/T - e-t/T)/(et/T + e-t/T)]

so if v = 0, then the the exponential functions on the numerator should equal 1 since -vt(1-1/1+1) = 0. This would be the case if the term (t/T) = 0 cause e0 = 1.

That's the farthest I got since I can't think of anything that can make the term equal zero unless ' t ' itself is zero. I think I'm missing something here.
 
  • #4
hotvette
Homework Helper
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I believe your expression for v(t) is incorrect, since v(0) should be v0.
 
Last edited:
  • #5
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I have the same question on my assignment. I don't get any hyperbolic functions because the object is moving upwards. It would become hyperbolic if it was moving downwards. The integral of 1/(1+x^2) is arctan. So instead of having tanh it would be changed to arctan.
 

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