# Can someone analytically solve this equation to confirm my Euler's Method answer?(Hard)?

I lack the current physics/math sophistication to solve this problem analytically, so I was instructed to use Euler's Method.

The upward flight of a model rocket can be approximated by the following equation

F-<m>g - rv^2 = <m>a

where <m> is the average mass of the rocket during the fuel burn and r is a resistive factor. The factor F is the thrust of the rocket engine which we assume is constant up burn out.

Assume <m> = 0.00792 kg, F = 4.17 N, and r = 0.000646 kg/m. The final mass of the rocket after burnout is 0.00751 kg. Let the time interval = 0.100 s and g = gravity = 9.81m/s^2.

If the engine burn time is 1.20 seconds, what is the speed and height at this time?

My answer: 44.9 m/s. The height is at 31.1 m.

How much higher does the rocket rise after the burn out?

My answer :The maximum height is 86.60 m at 4.30 seconds.

Determine the speed with which the rock hits theg round after its max height.

My answer: It hits the ground after the total time in the air is between 9.00 and 9.10, the speed is between 29.7 and 29.9 m/s.

So, I'm wondering if someone could tell me if I did this right? I rearranged the equation as follows:

a = F/<m> - g -rv^2/<m>

And then I solved for a0 and substituted that into v1, x1, a2, etc... using Euler's. At 1.20 seconds, I changed the equation in a11 to have no F force and changed the final mass.

Can someone solve this analytically to give me an approximate answer close to this, or maybe use Euler's Method themselves to confirm it? I *think* it's right.

(CROSS POSTED IN PHYSICS AND MATH)

Relevance

Rewrite the given force law in the form

(1) ..... a = F/<m> - g – r*(v^2)/ <m>

Substituting the given values in (1), we get the following

.......... a = 526.52 – 9.81 – (0.081566)*( v^2)

.......... a = 516.71 – (0.081566)*( v^2)

.......... a = - (0.081566)*[ ( v^2) - (516.71)/(0.081566)]

(2) ..... a = - (0.081566)*[ ( v^2) – 6,335 ]

Since a = dv/dt , we can rewrite (2) as

(3) ..... dv/dt = - (0.081566)*[ ( v^2) – 6,335 ]

Rewriting the term inside the bracket as the difference of two squares, which is

the product of a sum and a difference, (3) becomes

.......... dv/dt = - (0.081566)*[ v + 79.59 ]*[ v – 79.59 ]

(4) ..... dv/{ [ v + 79.59 ]*[ v – 79.59 ] } = - (0.081566)*dt

Rewriting the left hand side (LHS) of (4) as a sum of partial fractions of the form

.......... A/ [ v + 79.59 ] + B/[ v – 79.59 ]

where A and B are constants to be determined, we find that after taking the least

common denominator, we get the following:

.......... (A + B)*v = 0 so that B = -A

.......... A*(-79.59) + B*(79.59) = - 2*A*(79.59) = 1 so that

.......... A = -1/159.18 = - 0.006282 , B = +0.006282

It follows that (4) becomes

(5) ..... dv*{ (- 0.006282)/[ v + 79.59 ] + (0.006282)/[ v - 79.59 ] } = - (0.081566)*dt

Integrating the expression in (5), we get

.......... (- 0.006282)*ln [ v + 79.59 ] + (0.006282)*ln [ v - 79.59 ] = - (0.081566)*t + C1

.......... (- 0.006282)*ln [ v + 79.59 ] + (0.006282)*ln [ v - 79.59 ] = - (0.081566)*t + C1

.......... (0.006282)*ln [ ( v - 79.59 )/ ( v + 79.59 ) ] = - (0.081566)*t + C1

(6) ..... ln [(v - 79.59)/(v + 79.59)] = - (0.081566)*t/(0.006282) + C1 = - (12.984)*t + C1

where C1/(0.006282) is just another constant so we retained the C1.

Taking the exponential function of (6), we get

(7) ..... ( v - 79.59 )/ ( v + 79.59 ) = C1*exp [ - (12.984)*t ]

since exp(C1) is still a constant so we retained the C1. Assuming the model rocket

starts from rest so that v = 0 when t = 0, then (7) tells us that C1 = - 1. That means

.......... ( v - 79.59 )/ ( v + 79.59 ) = - exp [ - (12.984)*t ]

.......... v - 79.59 = - ( v + 79.59 ) *exp [ - (12.984)*t ]

.......... v*{ 1 + exp [ - (12.984)*t ] } = (79.59) *{ 1 - exp [ - (12.984)*t ] }

(8) ..... v = (79.59) *{ 1 - exp [ - (12.984)*t ] } / { 1 + exp [ - (12.984)*t ] }

Let θ = (12.984)*t . Then, we can rewrite the term inside the brace in (8) as follows:

.......... { 1 - exp [ - (12.984)*t ] } / { 1 + exp [ - (12.984)*t ] } = NUM/DEN

.......... NUM/DEN = [ 1 - exp ( - θ ) ] / [ 1 + exp ( - θ ) ]

.......... NUM = [ 1 - exp ( - θ ) ] = [exp ( θ/2)]*[ 1 - exp ( - θ ) ]

.......... NUM = [ exp ( θ/2) - exp ( - θ/2 ) ]

.......... DEN = [ 1 + exp ( - θ ) ] = [exp ( θ/2)]*[ 1 + exp ( - θ ) ]

.......... DEN = [ exp ( θ/2) + exp ( - θ/2 ) ]

.......... NUM/DEN = { [exp ( θ/2)]*[ 1 - exp ( - θ ) ] }/ { [exp ( θ/2)]*[ 1 + exp ( - θ ) ] }

.......... NUM/DEN = [ exp ( θ/2) - exp ( - θ/2 ) ] / [ exp ( θ/2) + exp ( - θ/2 ) ]

.......... NUM/DEN = sinh ( θ/2) / cosh ( θ/2) = tanh ( θ/2)

Since θ/2 = (12.984)*t/2 = (6.492)*t , we can rewrite (8) as

(9) ..... v = (79.59) *tanh [(6.492)*t ]

Substituting the engine burn time of 1.20 s, we get the velocity (speed):

(10) ... v = (79.59) *tanh [(6.492)*(1.20) ] = (79.59) *tanh [7.7904 ] = 79.59 m/s

To get the height at t = 1.20 s , we integrate (9) with respect to time:

.......... dy/dt = (79.59) *tanh [(6.492)*t ]

.......... y = (79.59)*integral { [ tanh (6.492*t ) ]*dt }

(11) .... y = [ (79.59)/(6.492)]*ln { cosh [(6.492*t )}

Substituting t = 1.20 s , we get

(12) .... y = [ (79.59)/(6.492)]*ln [ cosh ( 7.7904 ) ] = 87 m

But this is just for the first part of the motion , during the engine burn time only.

If you’re interested in knowing the rest of the analytic solution, you can e-mail me.

• Use algebra: Apply 3^x to both sides and obtain: 3^(x^2-13x+28) + 2/9 = 3^(log(base 2) 5) Subtract 2/9: 3^(x^2-13x+28) = -2/9 + 3^(log(base 2) 5) Take log(base 3) of both sides: (x^2-13x+28) = log(base 3) ( -2/9 + 3^(log(base 2) 5)) Subtract log(base 3) ( -2/9 + 3^(log(base 2) 5)) from both sides: x^2 -13x + 28 -log(base 3) ( -2/9 + 3^(log(base 2) 5)) = 0 You have yourself a nice quadratic equation. Apply the quadratic formula. The answer is messy, not particuarly enlightening but an analytical answer. The deep part of this is that you have to start treating things like log(base 3) ( -2/9 + 3^(log(base 2) 5)) like numbers because that's what they are, just plain old stupid numbers. The tricky part of math comes from something called "functional dependence". Even though your original equation looks messy as hell, it's still just some stupid quadratic.