/sci/ - Science & Math

v^2 - v0^2 = -2gh
set v = 0 at maximum, plug in g and h, solve fpr v0

>>9442893
wtf is this bait

>feet/sec
Fuck amerimutts.

>>9442919
wrong

>>9442893
>g
>ft/s

>>9442948
>wrong

>>9442893

>feet

I agree OP, most people on this board are not American

>>9442893
t=√(2h/g)
v=g*t

>>9442893
8 minutes per hour

>>9442893
>the velocity
>implying there is such a unique velocity

>not solving this in your head

>>9442893
>g=32ft/s
>s
>not squared
In my head I got like ~170 ft/s, but I am not used to dealing in retard units.

>>9443044
You, and you're shit hole country, have my pity.

>>9442978
If it points directly upwards (parallel to gravity force) then yes, there is an unique starting velocity.

>>9443249
>literally only country in the planet still dealing with imperial units because their population is too fucking retarded to make the change
>tfw

>>9443249
>56 IQ can't handle metric units
la tenebra......

>>9443305
>then yes, there is an unique starting velocity.
no, there isn't
learn to read, faggot

>>9442893
[math]\frac{1}{2}mv_0^2=\int_0^{h}dh'\frac{GmM_E}{(h'+R_E)^2}[/math]
[math]\frac{1}{2}v_0^2=\int_0^{h}dh'\frac{GM_E}{(h'+R_E)^2}[/math]
[math]\frac{1}{2}v_0^2=GM_E\int_0^{h}\frac{dh'}{(h'+R_E)^2}[/math]
[math]\frac{1}{2}v_0^2=GM_E\int_{R_E}^{h+R_E}\frac{du}{u^2}[/math]
[math]\frac{1}{2}v_0^2=GM_E\left.\frac{1}{u}\right|_{u=h+R_E}^{u=R_E}[/math]
[math]v_0=\sqrt{2GM_E\left(\frac{1}{R_E}-\frac{1}{h+R_E}\right)}[/math]

Plug in [math]h=1000[/math] ft and get the number yourself.

>feet
>feet/second

filthy american casual

>>9443351
Ah fuck, wasn't me.
[math]v_0=\sqrt{2GM_E\left(\frac{1}{R_E}-\frac{1}{h+R_E}\right)}[/math]

>feet

>>9443351

man you are autistic. this equation can be solved with a basic kinematics equation, not integral calculus. you are giving science a bad name. learn to recognize simple situations.

>>9443249
Dude I'm American and even I know acceleration in per unit time squared
get a fucking hobby, please

>>9443362

>>9443351
i really hate you "proves [math]\sqrt[3]{2} \notin \mathbb{Q} [/math] using fermats last theorem" type people, you find a way to give a correct answer while being utterly useless in helping someone understand a concept - and coming out smug about it no less

>>9442893
Depends on how long you want it to take to get that high. There's infidelity many solutions

>>9443368
>>9443437
sorry brainlets, KE = PE

>>9443532
tell me exactly why you even bothered answering

>>9443536
Obviously to brag to a science board that I know high school calculus.

>>9443544
it's sure what it looks like

>ft/s

>>9443351
why did you set kinetic energy equal to gravitational force integrated over height?

thats what you did right?

>>9442893
Am I still a brainlet if I could do this easily back in college but it's been years and I'd need to check up little-g...?

>>9443549
Good one.

>>9443570
Yup, the right hand side is just the potential energy change going from a height of 0 to 1000 ft. You then solve for the minimum velocity which will give you enough kinetic energy to reach that height.

>>9443532
>KE = PE
>K = P
there faggots solved it, where's my one million dollars now?

>>9442893
vf^2=vi^2+2*a*d

vf=0 at the apex, a=-9.81 m/s, d=304.8 m

vi=(2*9.81*304.8)^0.5 = 77.3 m/s

>>9443351
>Just a few more lines
>They're going to think I'm so smart
>They're going to post things like "Damn anon I wish I was as smart as you..."
>I'll feel really good about myself and happy with my life

>>9443630
>They're going to post things like "Damn anon I wish I was as smart as you..."
were*

>>9443635
>were
we'm'st'd've*

>>9443351
>not just using the conservation of energy

>>9443358
>this bad at MathJax
hello newfriend

>>9442893
wtf is this graph