Solving for T in a horizontal projectile equation

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SUMMARY

The discussion centers on solving for T in the horizontal projectile motion equation, specifically the equation 0 = Vi / K (1 - e^-kt)(sin theta) + (g / k^2)(1-kt-e^-kt). Participants highlight the complexity of the equation due to the presence of T both inside and outside the exponential function. A numerical solution approach is recommended, with suggestions to use the Lambert W function or Newton's method for faster convergence. The conversation emphasizes the iterative nature of solving such equations, particularly when dealing with air drag effects represented by the constant k.

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  • Knowledge of numerical methods for solving equations
  • Basic grasp of the Lambert W function and Newton's method
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Matt Jacques
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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
 
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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

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.
 
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.
 
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?
 
  • #10
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 I am stuck.
 

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