# 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...
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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)

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)

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