Getting acceleration due to gravity from dropping ball experiment This seems like pretty basic exp

ulcerat9jr

ulcerat9jr

Answered question

2022-05-02

Getting acceleration due to gravity from dropping ball experiment
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 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 t r a v e l e d = v i n i t i a l . t + 1 2 a t 2
a = 2 d t 2
a = 2.1 ( .4003 ) 2
a = 12.48 m s 2
2.
v f 2 = v i 2 + 2 a d
a = v f 2 v i 2 2 d
v f = d i s t a n c e t i m e = 1.0 0.4003 = 2.5 m s
a = ( 2.5 m s ) 2 2.1 m
a = 3.125 m s 2
3.
v f = v i + a t
a = v f v i t
a = 2.5 m / s 0 0.4003 s
a = 6.25 m s 2
I'm getting three different results. And all of them are far from 9.8 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?

Answer & Explanation

louran20z47

louran20z47

Beginner2022-05-03Added 14 answers

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, V e l o c i t y = D i s t a n c e T i m e , 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.
Cristian Rosales

Cristian Rosales

Beginner2022-05-04Added 26 answers

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.

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