V=v/t or square root of 2gh which one is the ideal speed and which one is the real speed in a projectile

Answer 1

As far as I know, we do not use "real speed" or "ideal speed" in physics BUT rather, average speed and or instantaneous speed.

Before I give the difference, let me explain some few terms.

1) Speed, v is the rate of change of distance, s with time, t. In symbolic form, v = s/t

a) supposing at some starting time, t1 and starting distance, s1, you moved to a final distance s2 with final time, t2, then your average speed is

= (s2 - s1)/(t2-t1) = Δs/Δt

b) In contrast, a car's speedometer shows its instantaneous speed, that is, the speed determined over a very small interval of time — an instant. Ideally this interval should be as close to zero as possible, but in reality we are limited by the sensitivity of our measuring devices. Mentally, however, it is possible imagine calculating average speed over ever smaller time intervals until we have effectively calculated instantaneous speed. In math form, this is written as v = ds/dt.

So now to answer your question. I must assume that your want to type v = s/t and NOT V = v/t, if I am correct, then the v = s/t is just the speed. However, when you are moving you have kinetic energy which comes from the mechanical energy in you (eg. from the delicious lunch you had -:) ) and this kinetic energy is transformed into potential energy supposing you are climbing up staircase (this is the origin of the height, h in your second equation)

Assuming energy is not lost in a form of heat/friction, then the kinetic energy is equal to the potential energy, therefore,

KE = PE ==> mv²/2 = mgh ===> v =√(2gh) . Since no energy is lost, you have 100% output (efficiency), this means that v =√(2gh) is the ideal speed and v = s/t is the real speed (in your case).