Approximating ballistic motion.

  • Thread starter Tschew
  • Start date
  • #1
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Hi!

I'm slightly stuck with an approximation of tmax, the time at which a ballistic projectile reaches maximum height. (projectile shot straigh upwards, no x motion)

So, by integrating the motion equation

m * dv/dt = -mg -bv

(where b is an air drag constant) I've created:

v(t) = (u + mg/b)*e^(-bt/m) - mg/b

which at tmax (t at maximum height) must be zero, which then gives:

tmax = (-m/b) ln [ (mg/b) / (mg/b + u) ]

where u is the initial velocity in y

Now when m >> b*tmax, tmax = u/g , howver, I can't get that!

all I get is (using e^x = 1 + x + x^2/2! +...):

tmax = u / (mgb/bm + ub/m)

thanks for any help!

-tschew
 

Answers and Replies

  • #2
ehild
Homework Helper
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Tschew said:
tmax = (-m/b) ln [ (mg/b) / (mg/b + u) ] *

where u is the initial velocity in y

Now when m >> b*tmax, tmax = u/g , howver, I can't get that!

all I get is (using e^x = 1 + x + x^2/2! +...):

-tschew


You can rewrite your equation labelled with "*" as

[tex] \exp(\frac{t_{max}b}{m})= 1+\frac{ub}{mg}[/tex]

As the exponent is much less than 1, you can apply the series expansion and stop at the linear term.

[tex] \exp(\frac{t_{max}b}{m})=1+\frac{t_{max}b}{m}[/tex]

[tex] 1+\frac{t_{max}b}{m} = 1 + \frac {ub}{mg}\rightarrow t_{max}=u/g
[/tex]

ehild
 
  • #3
11
0
Thank you.

Thanks for the reply. I noticed the minus sign in the ln() expression myself yesterday and got the result :eek:)

-Tschew
 

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