How much discrepency would I get if I used acceleration due to gravity?

[sodium40mg]

How much error would I get if I used 9.8 m/s^2 as the acceleration for a falling object in an experiment? For example, if I have a ball at, say, shoulder's length up in the air and I had someone measure the time it took for the ball to hit the ground. Would I get a lot of error if I used the equation

d = vi * t - 1/2 * a * t^2

Where vi = 0 m/s, so d = -1/2 * a * t^2 and a = - 9.8 m/s^2

Would that actually equal the distance from the shoulder length to the ground or how much error would I get? Would it be significant?

 

[Pengwuino]

Well exactly where are you trying to examine the erorr from? The fact that you're only using 2 significant figures?

 

[uart]

How much error would I get if I used 9.8 m/s^2 as the acceleration for a falling object in an experiment? For example, if I have a ball at, say, shoulder's length up in the air and I had someone measure the time it took for the ball to hit the ground. Would I get a lot of error if I used the equation


Would that actually equal the distance from the shoulder length to the ground or how much error would I get? Would it be significant?

Are you talking about measurement error in the timing?

Or are you talking more fundamentally about the applicability of the equation to the real world problem? The biggest discrepancy in the fundamental equation would be that it ignores air resistance. For a ball dropped from about head height the velocity is fairly low so for in most cases the effect of air resistance would be small, but if the ball was extremely low in mass then air resistance could significantly alter the result. In other words, the formula should be quite accurate for a cricket ball but not necessarily so for a ping pong ball.

 

[uart]

Here's a quick test of whether air resistance will be significant.

Final velocity (ideally) when dropped from 1.8m = 5.9 m/s

Air density at 20C, rho = 1.2 kg/m^2

1. Cricket ball. m = 0.16 kg, frontal area A = 0.0041 m^2, coeff of drag, cd = 0.47.

Air resistance at final velocity, F = 1/2 rho cd A v^2 = 0.041 N

Comparing the above with the force due to gravity of 1.6N we see that the drag is negligible.

2. Pin pong ball. Mass = 0.0027 kg, Frontal area A = 0.00126 m^2, drag coeff cd = 0.47

Air resistance at (ideal) final velocity, F = 1/2 rho cd A v^2 = 0.0125 N.

Comparing the above with the gravitational force of just 0.027 N we see that air resistance would be very significant in this case.

 

[PJC01]

I would like to clarify that the value of acceleration due to gravity, denoted as "g", is a constant and is approximately equal to 9.8 m/s^2 on Earth's surface. This means that if you were to conduct an experiment using this value, you would not get any discrepancy or error from the actual value.

However, if you were to measure the acceleration due to gravity in your experiment, you may get a slightly different value due to factors such as air resistance and the precision of your measuring equipment. This discrepancy would depend on the accuracy of your measurements and the conditions of your experiment.

In regards to the specific example you mentioned, using the equation d = vi * t - 1/2 * a * t^2 to calculate the distance a ball falls from shoulder length would yield the correct result. As long as you use the correct value for acceleration due to gravity, the equation will give you an accurate result.

It is important to note that in scientific experiments, it is always recommended to use precise and accurate measurements to minimize any potential errors. Any errors in measurements or calculations can lead to discrepancies in the results. Therefore, it is crucial to ensure that all variables, including acceleration due to gravity, are accurately measured and accounted for in the experiment.