- #1

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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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- Thread starter Matt Jacques
- Start date

In summary, the conversation is about solving an equation for T, but the equation is complicated due to the inclusion of the Lambert W function and air drag. Suggestions include using numerical methods or the Lambert W function, or solving iteratively using either Newton's method or picking a value for T and repeating the process. The conversation also involves clarifying the problem and attempting to simplify the equation.

- #1

- 81

- 0

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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- #2

- 1,605

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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...

- #3

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Yes, it is correct :(

- #4

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

- #5

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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.

- #6

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Newton's method will work faster.

- #7

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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)

- #8

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The equation is for air drag, notice the k.

- #9

- 81

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Ok, I got it down to

10^1+.5t = 19.6t

Can anyone help me on the Omega function?

10^1+.5t = 19.6t

Can anyone help me on the Omega function?

- #10

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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.

The horizontal projectile equation is a formula used to calculate the horizontal distance traveled by a projectile when launched at a certain angle and velocity. It is represented by the formula x = v₀t cosθ, where x is the horizontal distance, v₀ is the initial velocity, t is time, and θ is the angle of launch.

Solving for T in the horizontal projectile equation allows us to determine the time it takes for a projectile to travel a certain horizontal distance. This information is crucial in predicting the trajectory and impact of a projectile, making it important in fields such as physics, engineering, and ballistics.

To solve for T, you need to know the values of x, v₀, and θ. Simply plug these values into the equation x = v₀t cosθ and solve for t. You can use a scientific calculator or algebraic techniques to solve for t.

No, the horizontal projectile equation is only applicable for projectiles launched at a horizontal angle. For projectiles launched at an angle other than horizontal, other equations such as the vertical projectile equation or the general projectile motion equation must be used.

Yes, the horizontal projectile equation assumes that there is no air resistance and that the acceleration due to gravity is constant. In real-life scenarios, these assumptions may not hold true, leading to slight variations in the calculated horizontal distance.

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