This seems like pretty basic experiment, but I'm having a lot of trouble with it. Basically, I have two timer gates that measure time between two signals, and I drop metal ball between them. This way I'm getting distance traveled, and time. Ball is dropped from right above the first gate to make sure initial velocity is as small as possible (no way to make it 0 with this setup/timer). I'm assuming $v$ initial is $0 \frac{m}{s}$. Gates are $1$ meter apart.

Times are pretty consistent, and average result from dropping ball from $1.0$ meters is $0.4003$ seconds.

So now I have $3$ [constant acceleration] equations that I can use to get $g$.

  1. $$d_{traveled} = v_{initial} . t + \frac{1}{2} a t^2$$ $$a = \frac{2d}{t^2}$$ $$a = \frac{2.1}{(.4003)^2}$$ $$a = 12.48 \frac{m}{s^2}$$

  2. $$v_f^2 = v_i^2 + 2ad$$ $$a = \frac{v_f^2-v_i^2}{2d}$$ $$v_f = \frac{distance}{time}=\frac{1.0}{0.4003}=2.5 \frac{m}{s}$$ $$a = \frac{(2.5 \frac{m}{s})^2}{2.1 m}$$ $$a = 3.125 \frac{m}{s^2}$$

  3. $$v_f = v_i + at$$ $$a= \frac{v_f-v_i}{t}$$ $$a = \frac{2.5 m/s - 0}{ 0.4003 s}$$ $$a = 6.25 \frac{m}{s^2}$$

I'm getting three different results. And all of them are far from $9.8\frac{m}{s^2}$ . No idea what I'm doing wrong.

Also, if I would drop that ball from different heights, and plot distance-time graph, how can I get acceleration from that?


closed as off-topic by sammy gerbil, Bill N, Jon Custer, By Symmetry, Kyle Kanos Sep 7 '17 at 10:02

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" – sammy gerbil, Bill N, Jon Custer, By Symmetry, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Your methods #2 and #3 can't be used because you have no way of measuring the instantaneous final velocity with your setup. Also, your calculation for $v_f$ is really a calculation of the average velocity. As for Method #1, that can be applied but I'm not sure that putting a timer gate near the initial v=0 drop point is the best position to get the highest measurement accuracy of g. Seems to me that you might get more accurate g measurement by locating the first timer gate a bit further down (and, of course, accurately measuring its position). You should so an error analysis of your setup. $\endgroup$ – Samuel Weir Sep 5 '17 at 4:28
  • $\begingroup$ A big problem with your setup is that, as you are aware, you have no way of knowing what the true t=0 start time of the ball drop is. Putting the first timer gate as close as possible to the drop point doesn't seem like a good option to me because the ball will not have v=0 when it triggers the first timer gate and a small timing error can lead to a large error in $g$ (as Mitchell below showed). A work-around may be to instead put the timer gates at distances L1 and L2 down from the drop point and measure their trigger times t1 and t2. The L1=g (t1-t0)^2/2 and L2=g (t2-t0)^2/2, where t0 is .. $\endgroup$ – Samuel Weir Sep 5 '17 at 5:02
  • $\begingroup$ .. the (unknown) drop time. You then have two equations with two unknowns (i.e., 't0' and $g$). The unraveling of the solutions doesn't appear easy (I used Mathematica), but it is possible to get expressions for the actual start time 't0' and the acceleration $g$ from the measured quantities of L1, L2, t1, and t2. You just need to measure the distances L1 and L2 very accurately and hope that the gate timers are accurate and have little jitter. $\endgroup$ – Samuel Weir Sep 5 '17 at 5:06
  • $\begingroup$ Why use Mathematica? Set up a system of 3 equations and 3 unknowns: $v_1^2=2gL_1$, $v_2^2=2gL_2$, and $v_2=v_1+g\Delta t$. Solve for $g$. $\endgroup$ – Bill N Sep 5 '17 at 17:48
  • $\begingroup$ thanks guys, seems like yet again I started working without enough knowledge $\endgroup$ – Peter Sep 5 '17 at 21:06

The time you should be getting is $0.4516$ seconds. The measurement is off by $0.05$ seconds. This is reason why you are getting $12.48$ instead of $9.8$. This is one of the cases where even small errors in calculations can give you very wrong answers. Since the time is squared, it will bring more error to the answer.

Moving on, in your second and third calculations, you used a very wrong formula to get final velocity.

The relation,$Velocity=\frac{Distance}{Time}$, can only be used when motion in uniform (unaccelerated). But since the body is falling under gravity, the motion is accelerated.

Therefore, the last two calculations will always give wrong results because the usage of equations is wrong. However, the equations used in first equation are correct.


In #2 and in #3, You are calculating Vf as average velocity, not final velocity. With constant acceleration, your final velocity is twice the average so, it is 5 m/s, not 2.5 m/s. Now you will get all three results = ~ 12.5.

What this indicates is that your initial velocity is far from 0, and/or there is some other error in your experiment that you need to find yourself. It may be your time/distance measurements etc.


First of all the Physics.
The velocities in the constant acceleration kinematic equations are instantaneous velocities at the initial time equal to zero, $v_{\rm i}$, and and final time $t$ when the velocity if $v_{\rm f}$.

In the first equation that you used you made no use of the velocities other than stating that the initial velocity is zero and so that is the correct value of $g$ from your experimental data.

In the other two equations you assumed that the final velocity was $\frac d t$ which is in fact the average velocity assuming the ball started from rest.
For constant acceleration starting from rest the final velocity is twice the average velocity.
As the average velocity is $\frac {1.0}{0.4003} = 2.498\,\rm m/s$ then the final velocity is $2 \times 2.498 = 4.996 \,\rm m/s$.
This gives the acceleration of free fall $g = \frac {4.996}{0.4003}= 12.48\, \rm m/s^2$ as you found using your first equation.

Now on to the experiment which is difficult to comment on as the actual experimental set up has not be described in any detail.
Perhaps the first thing which need to be done is to get an estimate of what the "expected" time for a free fall of one metre using ?
This works out to be $\sqrt{\frac {2}{9.8}} \approx 0.45 \,\rm s$.
As the timing device itself seems to be of a high precision and probably of a reasonable accuracy the discrepancy of $10\%$ in the timing points to a flawed experimental arrangement/technique.

There are numerous ways of doing this experiment in the school/college laboratory but if you are able to drop the ball from rest from a fixed position I suggest that using your apparatus you try the following.

Have the first light gate a distance $D$ from the position where the ball is dropped and the second light gate at a position which is $D+d$ from the initial position of the ball.
Again measure the time interval between the ball passing through the first light gate and then through the second gate.
That time interval $t$ is the time it takes for the ball to travel, from an unknown but constant velocity $v_{\rm i}$ when at the first light gate, a distance $d$, to the second light gate.

$d = v_{\rm i} t + \frac 12 g t^2 \Rightarrow \frac d t = \frac 12 g \, t + v_{\rm i}$

Keeping the first light gate at the same position (ie keep $D$ constant) move the second light gate to a new position and hence a new value of $d$ and again measure the time interval.
You now have two equations and two unknowns $v_{\rm i}$ and $g$ and hence you can now solve for $g$.
Perhaps a better method of analysis might be to plot $\frac d t$, which is the average velocity between the two light gates, against $t$ and from the gradient of the graph ($=\frac g 2$) find $g$?

If you want any further help then perhaps you need to say more about the experimental set up that you used?

It may be of interest to you to you that your method is the basis of the method used by D R Tate to measure $g$ at the National Institute of Standards and Technology (NIST), then called the National Bureau of Standards. The paper is worth reading just to show what needed to be done to get a value to $\pm 0.00005\%$.

  • $\begingroup$ As the timing device itself seems to be of a high precision and probably of a reasonable accuracy... This is an unjustified assumption. The OP has not provided any description of the time measurement. The average time is quoted without any standard error. It may have been given a spurious level of accuracy. $\endgroup$ – sammy gerbil Sep 5 '17 at 11:49
  • $\begingroup$ @sammygerbil I agree with you comment in that there is no evidence about precision other than from the quoted time. However when light gates are used they are often connected to millisecond timers and this is what I have assumed to be true in this case. It would be hard to have the quoted time as the average of the time if it was a centisecond timer which was being used? $\endgroup$ – Farcher Sep 5 '17 at 11:51

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