View Full Version : Solving for T in a horizontal projectile equation
Matt Jacques
Apr12-03, 04:56 PM
Hi, I am stuck on even how to start to solve for T
the equations is:
0 = Vi / K (1 - e^-kt)(sin theta) + (g / k^2)(1-kt-e^-kt)
Any suggestions on how to begin to solve for T would be appreciated.
Thanks,
Matt
My oh my is this complicated....
Hmm... your problem is this bit:
1-kt-e^-kt
Since you have t both inside and outside the e^, I don't think there is an algebraic method for you to get an exact answer. Are you sure you formed the equation correctly?
I may be wrong though...
Matt Jacques
Apr12-03, 06:51 PM
Yes, it is correct :(
HallsofIvy
Apr13-03, 02:47 PM
Then you will need either to use a numerical form of solution or do a google search form "Lambert W function".
Couldn't you do it iteratively?
Pick a value for T and solve. Then take that result and plug it in for T and repeat.
If all goes well (depending on your pick to start...), it will converge on an answer.
HallsofIvy
Apr14-03, 11:45 AM
Pick a value of T and solve for what? [:)] What you are describing (I think!) is one very crude way of solving an equation iteratively.
Newton's method will work faster.
Sorry if this seems dumb ...
But how did the equation of an horizontal projectil get this complicated ?
The horizontal projectile equations that i know are well too easier ! (The one derived from the SUVAT equations)
Matt Jacques
Apr14-03, 02:12 PM
The equation is for air drag, notice the k.
Matt Jacques
Apr14-03, 08:58 PM
Ok, I got it down to
10^1+.5t = 19.6t
Can anyone help me on the Omega function?
Matt Jacques
Apr15-03, 08:40 AM
This is what I did:
I inserted some fixed constants and multiplied out
(48/.5)(1-^e-.5t)(sin 45) + (9.8/.25)(1-(.5t) - e^-.5t) = 0
(96 - 96e^-.5t)(sin 45) + 39.2(1-(.5t)-e^-.5t) = 0
67.88 - 67.88^e-.5t + 39.2 - 19.6t - 19.6e^-.5 = 0
107.2 - 87.48e^-.5t - 19.6t = 0
107.2 - 19.6t = 87.48^e-.5t
log(107.2 - 19.6t) = log(87.48^e-.5t)
log107.2 - log19.6t = -.5tLog(87.48)
2.030194 - log19.6t = -.5t(1.94198)
1.045463 - log19.6t = -.5t
-(1.045463 - log19.6t = -.5t)
-1.045463 + log19.6t = .5t
log19.6t = .5t + 1.045463
10^(.5t + 1.045463) = 19.6t
This is where Im stuck.
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