90% of Veeky Forums can't solve this on a piece of paper

>not solving this in your head

>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.

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

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

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

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

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

[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

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

>feet

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.

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

...

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

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

sorry brainlets, KE = PE

tell me exactly why you even bothered answering

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

it's sure what it looks like

>ft/s

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

thats what you did right?

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...?

Good one.

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.

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

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

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

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

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

>not just using the conservation of energy

>this bad at MathJax
hello newfriend

wtf is this graph