# Gravity in free fall equations

1. Sep 30, 2015

### jcharle111

Why is gravity negative in certain equations, such as
y = yi + vi t + (1/2) ( − g ) t^ 2 , but positive in others like
vf^2 =vi^2 +2gh ?

2. Sep 30, 2015

### tommyxu3

It should be correct that the second equation you wrote is $v_f^2=v_i^2+2(-g)h,$ where my $h$ means the displacement of the object.
If you write like that, you may regard $h$ as the height between $x_i$ and $x_f$, where $h$ is always positive.

3. Sep 30, 2015

### Ray Vickson

It depends on whether increasing values of y are upward (i.e., measured up from the ground) or are downward (i.e., measured down from a point above the ground, such as from the roof of a tall building). If increasing y is upward, the acceleration of gravity is in the opposite (downward) direction, so appears with a '-' sign. If increasing y points downward, the acceleration of gravity points in the direction of increasing y, so would appear with a '+' sign.

You can think about the second case yourself, but it essentially involves conservation of (total) energy.