0
$\begingroup$

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

$h = h_0 + v_0 t ± \frac{1}{2} g t^2$

and

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

Also, how are these called in English?

$\endgroup$
3
  • $\begingroup$ 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
    Commented Feb 4, 2013 at 23:49
  • $\begingroup$ They are called kinematics formulas. $\endgroup$ Commented Feb 4, 2013 at 23:53
  • $\begingroup$ They are also commonly called SUVAT equations - en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations $\endgroup$ Commented Feb 5, 2013 at 8:01

1 Answer 1

1
$\begingroup$

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 $v = u + at$.

$\endgroup$
7
  • $\begingroup$ h also could represent the height above a surface which is probably why g was still in it. $\endgroup$
    – tpg2114
    Commented Feb 5, 2013 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
    Commented Feb 5, 2013 at 1:53
  • $\begingroup$ based on what you have said, both formulas are identical. They both have the same name. $\endgroup$
    – Kenshin
    Commented Feb 5, 2013 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
    Commented Feb 5, 2013 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. $\endgroup$ Commented Feb 5, 2013 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.