I just want to know the differences between these two formulas:

$h = h_0 + v_0 t ± \frac{1}{2} g t^2$


$y = y_0 + v_{0y} t + \frac{1}{2} g t^2$

Also, how are these called in English?

  • $\begingroup$ The formulas are the same, just using a different variable to represent position. That is, y or h. My initial guess was that the top formula was horizontal, and the bottom was vertical, but if so, the top formula should have an A, instead of a g. $\endgroup$
    – Kenshin
    Feb 4 '13 at 23:49
  • $\begingroup$ They are called kinematics formulas. $\endgroup$ Feb 4 '13 at 23:53
  • $\begingroup$ They are also commonly called SUVAT equations - en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations $\endgroup$ Feb 5 '13 at 8:01

As far as I can tell, the two formulas look the same. In english, we write these formulas as:

In the horizontal (or x-direction):

$x = x_0 + v_{0x}t + \frac{1}{2}at^2$, where $x$ represents distance in the horizontal direction, $x_0$ represents the initial distance, $v_{0x}$ is the initial velocity in the x-direction (sometimes instead written as $u_x$) and finally $a$ is the acceleration in the horizontal direction.

Similarly, we have for the vertical, or y-direction:

$y = y_0 + v_{0y}t + \frac{1}{2}gt^2$

Notice I have replaced $a$ with $g$, since in most problems, the acceleration in the y-direction is due to gravity, so $a=g$.

These equations are referred to as "kinematic equations" along with these additional formulas:

$v^2 = u^2 + 2as$ and $v = u + at$.

  • $\begingroup$ h also could represent the height above a surface which is probably why g was still in it. $\endgroup$
    – tpg2114
    Feb 5 '13 at 1:08
  • $\begingroup$ Yes, h represents the height. Also, when I asked the names I meant the formula's name, I guess the first one is "vertical position as a function of time", but I don't know the second, could it be the same? $\endgroup$
    – Adami
    Feb 5 '13 at 1:53
  • $\begingroup$ based on what you have said, both formulas are identical. They both have the same name. $\endgroup$
    – Kenshin
    Feb 5 '13 at 2:27
  • $\begingroup$ @Chris so why in the first one I can user positive or negative gravity (pointing up or down) and in the other just positive? $\endgroup$
    – Adami
    Feb 5 '13 at 11:03
  • 1
    $\begingroup$ @Adami: No, the direction of g does not depend on the direction your ball is moving. It only depends on the orientation of your chosen set of coordinate axes. $\endgroup$ Feb 5 '13 at 18:27

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