Solving for T in a horizontal projectile equation

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
 

FZ+

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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...
 
Yes, it is correct :(
 

HallsofIvy

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HallsofIvy

Then you will need either to use a numerical form of solution or do a google search form "Lambert W function".
 

enigma

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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

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Pick a value of T and solve for what? :smile: What you are describing (I think!) is one very crude way of solving an equation iteratively.

Newton's method will work faster.
 
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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)
 
The equation is for air drag, notice the k.
 
Ok, I got it down to

10^1+.5t = 19.6t

Can anyone help me on the Omega function?
 
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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