What power motor should be used to power a Segway-like device
Too wide a question set, and some don't matter.

How fast?:  30 kph is getting fast for something small and unstable. Energy at 30 kph is \$(\dfrac{30}{20})^2\$ which is over double of that at 20 kph. Probability of pedestrian death in vehicle impact is \$\sim\sim\sim\sim\dfrac{V^2}{5000}\$ (V in kph). At 30 kph, the chance is \$\dfrac{30^2}{5000} = \sim 20%\$. That formula is for cars and pedestrians but gives you some indication. 20 kph is a safer target for playing an takes less than 1/3 of the windage power that 30 kph does ! - see below.



Battery is probably lead acid as available cheap per energy content or surplus. Low energy per mass and energy per size are tolerable.

Voltage probably 12V or 24V. One or two x 12V batteries. More is possible but gets bulky / annoying. 12V is good enough. 24V favoured by commercial designs as lower current so lower IR wiring losses.

12V or 24V DC motors in the hundreds of Watts range are available on the surplus market BUT are not usually cheap as they ar sought after by people who want to do similar to what you wish to do - vehicles, robots, ... . 



Windage = power losses due to air drag.


Empirical formula - power required to overcome windage for a compact one person vehicle is 


$$ Windage \, power = \frac{V^3}{180} $$

Power in Watts. V in kilometres per hour. Examples:


160 kph gives 22,500 Watts  
30 kph gives 150 Watts  
20 kph gives 44 Watts.


This is simply a translation of the old motorcyclists adage   


  It takes 30 horsepower to ton for a well tucked in rider with leathers :-)


There are other ways of calculating this, but that is good enough in such an inexact area. For example:

$$Power = 0.5 \times air \, density \times frontal \, area \times drag \, coefficient \times velocity^2$$

The motorcycle formulae is as easy :-).  

For something approximating a flat plate this gives 

$$\sim\sim\sim\sim Power = A \times V^3 $$ 

A in \$m^2\$, V in \$m/s\$



Take off one V and you get Drag (Newtons) \$\sim\sim A \times V^2\$.

This works well enough for bowling balls, raindrops, skydivers, field mice and parachutes. 



SO

ie you need a say 200 Watt motor at minimum to get something like your desired 30 km/h top speed on the flat. Using a 12V battery that's about 16 amps. 

An eg 7Ah "brick" alarm system battery would wilt very quickly under that current. Notionally 7/16 hour = 26 minutes but really quite a bit less.

A say 30 Ah car battery will notionally give you approaching two hours at top speed. Somewhat less in practice.

Look at the many small electric scooters/bikes/xxx around and see what they use and what ranges and speeds they claim.



Uphill

$$Power = \dfrac{Newton \times metres}{second} \text{or} \sim \dfrac{kg \times m}{s} \times 10 $$

To ascent a slope at 100% efficiency you need  

$$ \sim Power = \frac{Mass \times height \, change}{second} \times 10 Watts $$

Mass in kg, \$h\$ is increase in vertical height per second.

To lift you up the slope in addition to windage. 

For example, at 30 kph ~= 8 m/s if you ascend a 1 in 20 slope (about 3 degrees) your vertical height change per second is ~= 8/20 = 0.4 ms.

If all up vehicle and rider weight is 100 kg, then

$$Power \, needed = M \times h/s \times 10 = 100 \times 0.4 \times 10 = 400 Watts$$

This dominates your windage losses on hills of this slope.