Solving for T in a horizontal projectile equation

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Discussion Overview

The discussion revolves around solving for the variable T in a complex equation related to horizontal projectile motion, specifically incorporating air drag effects. Participants explore various methods for approaching the problem, including numerical solutions and iterative techniques.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Matt expresses difficulty in starting to solve the equation involving T and asks for suggestions.
  • One participant questions the algebraic solvability of the term "1-kt-e^-kt" due to the presence of T both inside and outside the exponential function.
  • Another participant suggests using a numerical solution or the Lambert W function as potential methods to solve the equation.
  • An alternative approach of iterative solving is proposed, where a value for T is chosen, solved, and then refined through repeated substitutions.
  • A participant mentions that Newton's method could provide a faster iterative solution compared to the crude method initially described.
  • Matt questions the complexity of the equation compared to simpler horizontal projectile equations derived from SUVAT equations.
  • Another participant clarifies that the complexity arises from the inclusion of air drag, indicated by the variable k.
  • Matt shares progress in simplifying the equation down to a logarithmic form but indicates being stuck at a particular step involving the Omega function.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to solve for T, with multiple competing approaches and uncertainties expressed regarding the complexity of the equation.

Contextual Notes

Participants note the presence of air drag in the equation, which complicates the traditional horizontal projectile motion equations. There are unresolved mathematical steps in the simplification process, particularly regarding the logarithmic transformation and the Omega function.

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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