Air Resistance and Projectile Motion

AI Thread Summary
The discussion focuses on calculating the time to reach maximal elevation for an object ejected straight up, considering both gravity alone and gravity with a retarding force. For part (a), the time is determined as tmax = v0/g, while part (b) involves a second-order differential equation yielding tmax = (2/k) ln([kv0/g]-1). The user encounters difficulties in expanding the result from part (b) to match part (a) as k approaches zero, initially struggling with a 0/0 form. After considering l'Hôpital's rule, they successfully simplify the expression and correct an earlier algebraic mistake. The conversation emphasizes the importance of correctly applying mathematical techniques to solve physics problems involving projectile motion and air resistance.
AlexChandler
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Homework Statement



An object is ejected straight up into the air at an initial velocity v0.
(a) Determine the time for reaching the maximal elevation when the object is subject
to gravity alone.
(b) Determine the time for reaching the maximal elevation when the object is subject
to gravity combined with a retarding force of the form kmv.
(c) Carefully expand your result from 1b to determine that it agrees with 1a in the
limit of k approaching 0.
(d) On the basis of the expansion 1c decide whether the retarding force extends or
shortens the time to reach the maximal elevation.

Homework Equations



Fz=maz
Fz=-mg
Fz=-mkv

The Attempt at a Solution



I had no real trouble with part a and b

For a I got tmax=v0/g

for b I had to use a 2nd order Diff Eq and I got tmax=(2/k) ln([kv0/g]-1)

for part c I tried using a Maclaurin polynomial for my result t(k)...(t as a function of k) for part b around k=0. However I was unable to do the expansion as I could not find the value of t(0)... (this is the value for the time as calculated in part b as k approaches zero) I tried using a limit to evaluate the value, but couldn't figure it out. Any ideas?
Thanks
 
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If you put k=0, you will get a 0/0 form,
So, you can apply L' Hospital Rule!
 
Yes, I thought of that at first, but actually you get ln(-1)/0
and natural log(-1) = 3.14159265 i
 
AlexChandler said:
for b I had to use a 2nd order Diff Eq and I got tmax=(2/k) ln([kv0/g]-1)

Show your work. You have to solve a first-order equation to get the time of maximal elevation.

ehild
 
F=ma. In vertical dimention (z) with up as positive
-kmv-mg=ma
-kmz'-mg=mz"
z"+kz'= -g
this is the second order diff eq I solved to find a function z(t)
I then set z'(t)=0 and solved for the time t at which z is max
this led me to the solution I posted.
 
Haha I suppose it would be first order.
v'+kv=-g
Since I don't even really need the position function.
Is this the first order you were talking about?
 
Ahh yes have solved the problem. Made a small algebraic mistake a few steps ago. The formula for Tmax turned out to be much simpler to expand. Actually, I was able to use l'Hôpital's rule! (that is... Bernoulli's rule). Thanks for your help.
 
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