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The formula for air resistance is $$F_{air}=\frac{1}{2}\rho cAv^2 $$

where $\rho$ is the density of the air, $c$ is the drag coefficient, $A$ is the cross sectional area of the object and $v$ is the speed of the object.

We see that air resistance is independent of mass. So two indentically formed objects (such as a bowling ball and a basket ball) will hit the ground at the same time, if they are both released from the same height and at same time.

However, while the object falls, its speed increases until the air resistance equals the gravitational force on the object. When that happens, the object has reached terminal velocity.

$$F_g=F_{air} $$ $$mg=\frac{1}{2}\rho cAv_{\text{terminal}}^2 $$ $$v_{\text{terminal}}=\sqrt{\frac{2mg}{\rho cA}} $$

We see here that the terminal velocity is dependent on the mass. Meaning that a bowling ball will reach a much greater speed during the fall, than a basket ball.

My question

How is it possible for a bowling ball and a basket ball to hit the ground at the same time?

The bowling ball reaches a much larger travelling speed (terminal velocity) than the basket ball and they both encounter the same air resistance. For me, this is an indication that the bowling ball hits the ground first, but this (apparently) is not the case.

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How is it possible for a bowling ball and a basket ball to hit the ground at the same time?

It isn't. Unless it is in a vacuum (no air).

Both objects continue to accelerate until they reach their terminal velocity. The object with a higher mass has a higher terminal velocity (the bowling ball) and will hit the ground first because it continues to accelerate after the object with the lower mass and lower terminal velocity (the basket ball) stops accelerating. So for the fall, the average acceleration of the heavier object is greater.

The bowling ball reaches a much larger travelling speed (terminal velocity) than the basket ball and they both encounter the same air resistance.

Yes, and that's why the bowling ball will hit the ground sooner than the basket ball. The upward force of air resistance is the same on both, but the downward force of gravity, which is $mg$, is greater for the object with greater mass. So for the fall, the average acceleration of the heavier object is greater.

$$h=\frac{a_{ave}t^2}{2}$$

$$t=\sqrt{\frac{2h}{a_{ave}}}$$

Since the average acceleration of the heavier object is greater than the lighter object, the time $t$ it takes to reach the ground is less.

Hope this helps.

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In air the acceleration depends on the mass, so the objects will not reach the ground at the same time.

The sum of the forces with air resistance.

$m a = m g-\frac{1}{2} A \text{$\rho $c} v^2$

or

$a=g-\frac{A \text{$\rho $c} v^2}{2 m}$

So that the acceleration $a$ indeed depends on the mass. Without air resistance $a=g$ and the acceleration is independent of the mass.

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