Really basic question, but basically I'm given a change in height in centimeters (that's how I measured it). From that, I'm supposed to find the initial kinetic energy.

$KE_i = PE_f$

After doing some work, the masses cancel out and I'm left with:

$v_i=\sqrt{2g\bigtriangleup h}$

So my change in height is 8.80 cm. If I input that, my velocity turns out to be 13.1 cm/s. However, if I initially convert it to meters, it ends up being 1.31 m/s. 13.1 centimeters definitely isn't the same as 1.31 meters.

I can see that the square root is the problem, but why? And what should I actually do?

In addition, if there is a square root in kinematic equations as well, does it need to be in meters or something?


closed as off-topic by Bill N, user36790, CuriousOne, AccidentalFourierTransform, Qmechanic May 5 '16 at 10:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Check your units again, very carefully. What value of $g$ did you use to get 13.1 cm/s? $\endgroup$ – tpg2114 May 5 '16 at 2:55
  • $\begingroup$ If you are using cm for the change in height what units are you using for g? They should be 980 cm/s/s if you want to get speed in cm/s. $\endgroup$ – M. Enns May 5 '16 at 2:56
  • $\begingroup$ Did you use 980 cm/s^2 for g? Always check for consistent units! $\endgroup$ – Bill N May 5 '16 at 2:56
  • 4
    $\begingroup$ I'm voting to close this question as off-topic because the OP didn't check for proper units. $\endgroup$ – Bill N May 5 '16 at 2:57

I think your problem is that you didn't change the units in the constant g. It has a value of approximately $9.8ms^{-2}$. Notice that it depends on meters. To obtain the correct result, you should use $980cms^{-2}$. Notice that this constant is off by a factor of 100, so that the result (after the square root) is off by a factor of $\sqrt{100}=10$.


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