Equation for height as a function of time under gravity

[nhmllr]

If I were to create a function for height affected by gravity (as a function of time) would it be
height = initial upwards velocity * time - gravity * time²
OR
height = initial upwards velocity * time - 2 * gravity * time²
OR something else, and why?
I thought it would be the first one, but after trying to solve some problems, upon checking my work it seems like it would be the second one.

 

[Doc Al]

Neither of those are correct. The correct expression would be:
height = initial upwards velocity * time - 0.5 * gravity * time²

(Review the https://www.physicsforums.com/showpost.php?p=905663&postcount=2".)

 

[daumphys]

Good answer. But it has small differences.
height = initial height + initial upward velocity * time - 0.5* gravity * time^2
where moving direction is the counter gravity direction.

 

[nhmllr]

Neither of those are correct. The correct expression would be:
height = initial upwards velocity * time - 0.5 * gravity * time²

(Review the https://www.physicsforums.com/showpost.php?p=905663&postcount=2".)

Ah. After reviewing my work, it turns out that I should have meant to divide by two instead of multiply.
In anycase, what is the reason behind that 1/2? What's it doing that makes it right? It probably has something to do with integrals, because it's in the t² term, but what?

 

[Doc Al]

It probably has something to do with integrals, because it's in the t² term, but what?

The acceleration is constant. Integrating once gives the velocity:
v = v₀ + at

A second integration gives the position:
x = x₀ + v₀t + 1/2at²

Setting x = height, x0 = 0, and a = -g, gives the equation you need:
h = v₀t - 1/2gt²

 

[nhmllr]

The acceleration is constant. Integrating once gives the velocity:
v = v₀ + at

A second integration gives the position:
x = x₀ + v₀t + 1/2at²

Setting x = height, x0 = 0, and a = -g, gives the equation you need:
h = v₀t - 1/2gt²

Thanks, I get it now