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KE

Hi all,

I have again decided to build a smallish rocket to ride around in. I plan to truck along about 100 feet above the flat sand territory of Nevada where the air resistance will be constant. Also the rocket mass (including me, of course) and power I have, along with gravity will be constant. I plan to go fairly fast, but not too fast note, i.e. accelerate in a sinusoidal fashion up to 500 mph and lope along at that speed. Now obviously the kinetic energy of the rocket and myself will be equal to the energy expended (work done) to get me up to speed, however energy will be lost to friction due to the air during acceleration. My concern is how to calculate the kinetic energy the rocket and myself will have at the moment I arrive at 500 mph. Anyone care to offer a formula for this?

jc the elder

Reply to
jclause
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are you implying that the wind never blows in nevada?

how will you crerate sinusoidal acceleration with a constant thrust engine?

Now obviously the

first calculate the ballistics of the rocket to determine how many seconds until you crash.

Reply to
TimPerry

you do know the formula for KE, don't you? It has something to do with half of the square of velocity and mass, It has nothing to do with how you got to that velocity.

Reply to
Tony

Reply to
no_one

So you have invented a mass-less fuel to use? From my physics and calculus days I seem to remember (more like a nightmare) hating rocket problems because the mass was constantly changing.

Sounds to me like you plan on getting a "Darwin Award" very fast.

Chalres Perry P.E.

Reply to
Charles Perry

Let's assume a quiet day with no wind.

My fuel injection feeds the flow in a sinusoidal fashion, note.

I plan a soft landing, but need to know the KE upon arriving at 500 mph.

jc the elder

Reply to
jclause

No, it is (1/2 M vfinal^2) - (1/2 M vinitial^2) , not using just v. Try (in this case, starting from standstill) KE = v^2 M.

jc the elder

Reply to
jclause

KE=1/2 M vfinal^2, if starting from standstill.

Reply to
jclause

Further thoughts, eh. Very good, you are correct. OK, let's assume the mass of the fuel averages .707 its initial value, and go with that. BTW, I plan to hold my own mass constant (no beans before takeoff) :)

jc the elder

Reply to
jclause

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