Having trouble with this kinematic question as I'm just learning physics again, how exactly do I start this question.
A rocket weighing 1104672 lbs is launching using a 3 million lb constant-thrust engine. Calculate the rocket's theoretical velocity at a height of 1 km.
Knowing that (velocity = acceleration * time) and (force = mass * acceleration, calculate how long the rocket takes to reach 1km high.
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You need to know how much fuel and oxidizer is consumed by the rocket's engines per second to answer that.
Forgot to mention the question is asking to find km/hr and how long it takes in seconds
I'm sure the problem was just written as a simple example without considering of the realities of rocketry
Solve for time first using distance and acceleration, then find velocity using time and acceleration
I'm assuming I have to convert km to feet? Also can I divide the weight by 32.2 slugs in order to find mass?
well you know the acceleration because its 3 million / 1.1 - 1, gives 1.7 g's of acceleration
so you just kinda do the rest...
Yes! You also have to convert Imperial seconds to Metric seconds
why the devil would u need to convert to feet
do not forget you must take into account gravity of the earth, but also gravity of other planets in the solar system pulling on the rocket.. the question didn't disclose the date and time? because you can't solve this problem at all unless you know where the other planets are in orbit
Height is in metric while weight is in imperial.
so you have to convert time to imperial as well
[math]
h=h_0+v_0t+\frac{1}{2}at^2\Leftrightarrow t = \frac{-v_0 \pm \sqrt{v_0^2 - 4\cdot \frac{1}{2}a\cdot (h_0-h)}}{2\cdot \frac{1}{2}a}
\\
F=ma \Leftrightarrow a=\frac{F}{m}
[/math]
>A rocket weighing 1104672 lbs is launching using a 3 million lb constant-thrust engine. Calculate the rocket's theoretical velocity at a height of 1 km.
Well, dividing 1104672 lbs by 3 million lb gives 0.368 s, which can only be the time.
Then 3 million / 1104672 gives you the acceleration: 2.72. Now you just plug acceleration * time: 2.72*0.386=1.05 km/s.
Depends on how fast the fuel is being consumed
What is the coefficient of drag and how many astronuts are on board
>weighing 1104672 lbs is launching using a 3 million lb constant-thrust engine
• Step One: stop using antiquated "lb" units
• Step Two: Lrn2rocketry fgt pls
1) convert lbs to newtons and stay in metric the rest of the way. I'm not going to do that because it's your homework, Both thrust and weight are forces. Solving for mass is irrelevant.
2) Problem can't be solved as stated. Rockets lose mass continually so the acceleration varies unless the engine is throttled. Which it isn't in this case.
Assuming it's just a poorly written problem and acceleration is constant, the thrust in excess of it's weight is 3,000,000 - 1,104,672 = 1,895,328. Effective acceleration is 1.716 gravities. (The passengers feel 2.716 The difference is thrust wasted in just "suspending" the rocket.)
Convert the acceleration into meters/sec2 and then D=.5 A T^2 and V = A T
>Rockets lose mass continually
how much mass does falcon9 spit out as a function of time, or height?
tried to google it but didn't find anything useful
Rockets are rated by specific impulse, Isp.
It has the units of seconds.
1 lb of fuel produces a thrust of 1 lb for a duration of N seconds.
Alternatively, burning a lb of fuel in 1 second produces N lbs of thrust.
en.wikipedia.org
Isp seems about 311.
If it produces 155,000 lbs force, it's swallowing 155,000/311 = 498 lbs/sec.
The Falcon 9 has 9 of these in the first stage.
You'll note slightly different Isp given for sea level and vacuum. Engine efficiency increases a bit if there's no air to get in the way.
>498 lbs/sec. x9
roughly 2 metric tons per second
in youtu.be
takes 25sek to get to 1 km
so, loses 50 tons in that time
thanks mate
Don't forget that it's not constant acceleration.
When rockets are fully loaded, they just crawl upwards at the start, wasting most of the thrust in simply opposing gravity.
That's why most designs have solid boosters.
Solids have lower Isp but produce a lot of thrust for a short period. They get you past those inefficient first seconds.
To get the actual time if acceleration varies you have to either solve a differential equation or integrate numerically, say at 1 second interval. Excel can do the latter easily.
>Don't forget that it's not constant acceleration.
I know, that's why I cheated by looking at the video :)
(btw I'm not OP)