And thus we get back to the question of what’s important in physics equations.
But let’s do a numerical example for fun.
Our ball is 5 cm in diameter, so its volume is about 65.5 cm3. Let’s make it out of wood, say, bamboo. Its density is about 0.35 g/cm3 so the ball will weigh about 23 g.
Let’s calculate its terminal velocity, that is, the speed at which drag exactly balances gravity. The formula is v = sqrt(2mg/(pAC)) where m is mass (0.023 kg) , g is the same old 9.8, p is air density which is about 1.2 kg/m3, A is projected area and since we have a sphere it’s 19.6 cm2 or 0.00196 m2, and C is the drag coefficient which for a sphere is 0.47.
So the terminal velocity of a 5 cm diameter bamboo ball is about 20 m/s. That is quite a way off your estimate of 34.3 and we got there without using things like hollow balls or aerogel :-)
To be fair, a light ball is exactly where my estimate is known to be least accurate. Let’s consider, rather, a ball with a density of 1 - one that neither floats nor sinks in water. (Since, in my experience, many things sink in water and many, but not quite as many, things float in it, I think it makes a reasonable guess for the average density of all possible balls). Then you have m=0.0655kg, and thus:
...okay, if it was falling in a vacuum it would have reached that speed, but it’s had air resistance all the way down, so it’s probably not even close to that. (And it it had been dropped from, say, 240m, then I would have calculated a value of close on 70 m/s, which would have been even more wildly out).
So, I will admit, it turns out that mass is a good deal more important than I had expected—also, air resistance has a larger effect than I had anticipated.
And thus we get back to the question of what’s important in physics equations.
But let’s do a numerical example for fun.
Our ball is 5 cm in diameter, so its volume is about 65.5 cm3. Let’s make it out of wood, say, bamboo. Its density is about 0.35 g/cm3 so the ball will weigh about 23 g.
Let’s calculate its terminal velocity, that is, the speed at which drag exactly balances gravity. The formula is v = sqrt(2mg/(pAC)) where m is mass (0.023 kg) , g is the same old 9.8, p is air density which is about 1.2 kg/m3, A is projected area and since we have a sphere it’s 19.6 cm2 or 0.00196 m2, and C is the drag coefficient which for a sphere is 0.47.
v = sqrt( 2 0.023 9.8 / (1.2 0.00196 0.47)) = 20.2 m/s
So the terminal velocity of a 5 cm diameter bamboo ball is about 20 m/s. That is quite a way off your estimate of 34.3 and we got there without using things like hollow balls or aerogel :-)
To be fair, a light ball is exactly where my estimate is known to be least accurate. Let’s consider, rather, a ball with a density of 1 - one that neither floats nor sinks in water. (Since, in my experience, many things sink in water and many, but not quite as many, things float in it, I think it makes a reasonable guess for the average density of all possible balls). Then you have m=0.0655kg, and thus:
v = sqrt( 2 0.0655 9.8 / (1.2 0.00196 0.47)) = 34.0785 m/s
...okay, if it was falling in a vacuum it would have reached that speed, but it’s had air resistance all the way down, so it’s probably not even close to that. (And it it had been dropped from, say, 240m, then I would have calculated a value of close on 70 m/s, which would have been even more wildly out).
So, I will admit, it turns out that mass is a good deal more important than I had expected—also, air resistance has a larger effect than I had anticipated.