# If I dropped a golf ball straight down from about 5 meters, would air resistance cause a noticable change in the acceleration of the ball?

I am working on a Grade 11 physics lab report. I dropped a golf ball from a balcony 5m and 13cm off the ground. When I did all the calculations, the acceleration was 7,72m/s (calculations below). I am wondering if this discrepancy (7,72m/s² < 9,8m/s²) was caused by the human reaction time being slow, by air resistance or by human error (a miscalculation somewhere).

Calculations (v = velocity):

average ∆t = 1,15s, ∆d = (-)5,13m, v = ?, vi = 0m/s, a = ?

average v = ∆d/∆t
= 5,13m/1.15s
= 4,46m/s

final v = 2(average v) - initial v
= 2(4,46m/s) - 0m/s
= 8,92m/s

a = (final v - initial v)/∆t
= (8,92m/s - 0m/s)/1,15s
= 7,76m/s²


You are on the right path in terms of trying to run down possible sources of error.

You can estimate the drag force on the ball as a function of speed $$v$$.

Looking at https://core.ac.uk/download/pdf/82404896.pdf I see that the coefficient of drag for a golf ball at low speeds (low Re number) is about $$C_D = 0.65$$ for typical balls.

So let's calculate the deceleration due to drag at $$v=8.92\;\text{m/s}$$

Force of air resistance $$F_{\rm drag} = \tfrac{1}{2}\rho C_D A v^2 = 0.044\; \text{N}$$

where $$A=0.001432\;\text{m}^2$$ is the frontal area of a ball, and $$\rho = 1.2\;\text{kg/m}^3$$ is the density of air

Acceleration due to drag

$$a_{\rm drag} = \frac{F_{\rm drag}}{m} = 0.968\;\text{m/s}^2$$

where $$m=0.046\;\text{kg}$$ is the mass of the ball.

As you can see the deceleration due to drag is about 10% of gravity in the end, and thus you cannot ignore its effects.

But this 10% effect, does not account for the 21% change in apparent acceleration.

So the most likely scenario here is measurement error

If you learn calculus in the future you will learn how to deal with problems with variable acceleration. I did some quick calculations with a first-order approximation and you can estimate the measured acceleration $$g'$$ as a function of earth's $$g$$ and the distance traveled as $$g' = g \left( \tfrac{5}{3} - \tfrac{2}{3} \exp(\beta \Delta d/m) \right)$$

where $$\beta = \tfrac{1}{2}\rho C_D A = 0.000558\;\text{kg/m}$$, $$\Delta d = 5.13\;\text{m}$$ is the distance traveled and $$g=9.81\;\text{m/s}^2$$

With the above values, if you had a perfect timing system, you would have calculated $$g' = (95.8\%) g = 9.389 \;\text{m/s}^2$$ instead of 7.72

Doing the math with $$g=9.81\;\text{m/s}^2$$ I see the ball will need $$\Delta t = 1.0333\;\text{s}$$ to reach $$v_f = 9.728\;\text{m/s}$$ and $$\Delta d= 5.13\,\text{m}$$. This means your timing error is about $$1.15 - 1.033 = 0.1167\,\text{s}$$ too slow.

It's most likely reaction time. Air resistance will be a small effect in a 5 meter drop. To get accurate results in an experiment like this, you need some mechanism to trip the clock at the beginning of the fall and at its end.

Another bit of advice: it's always best to repeat the experiment 10 times and take averages.