Why the large ball falls farther than the smaller one?

[Chestermiller]

The actual force balance equation is $$ma=mg\left(1-\frac{\rho_a}{\rho}\right)-\rho_a\frac{v^2}{2}\pi R^2C_D$$where $\rho_a$ is the density of air, $\rho$ is the density of the ball material, v is the falling velocity of the ball, R is the radius of the ball, and $C_D$ is the air drag coefficient (assumed constant for the present development). The first term on the right hand side is the net downward force of gravity and buoyancy. The second term on the right hand side is the air resistance force.

If we divide both sides of this equation by the mass of the ball, we have
$$a=g\left(1-\frac{\rho_a}{\rho}\right)-\rho_a\frac{v^2}{2}\left(\frac{\pi R^2}{m}\right)C_D$$Note that the first term on the right hand side of this equation is independent of the mass of the ball. But the second term is not. The term $\left(\frac{\pi R^2}{m}\right)$ is proportional to the surface area to mass ratio of the ball. Since the mass of the ball is equal to its volume $\frac{4}{3}\pi R^3$ times its density $\rho$, we have that $$\left(\frac{\pi R^2}{m}\right)=\frac{3}{4\rho R}$$Since R is smaller for the smaller ball than for the larger ball, the expression is larger for the smaller ball than for the larger ball. That means the the effect of air resistance on the ball acceleration is greater for the smaller ball than the larger ball.

 

[YMMMA]

The actual force balance equation is $$ma=mg\left(1-\frac{\rho_a}{\rho}\right)-\rho_a\frac{v^2}{2}\pi R^2C_D$$where $\rho_a$ is the density of air, $\rho$ is the density of the ball material, v is the falling velocity of the ball, R is the radius of the ball, and $C_D$ is the air drag coefficient (assumed constant for the present development). The first term on the right hand side is the net downward force of gravity and buoyancy. The second term on the right hand side is the air resistance force.

If we divide both sides of this equation by the mass of the ball, we have
$$a=g\left(1-\frac{\rho_a}{\rho}\right)-\rho_a\frac{v^2}{2}\left(\frac{\pi R^2}{m}\right)C_D$$Note that the first term on the right hand side of this equation is independent of the mass of the ball. But the second term is not. The term $\left(\frac{\pi R^2}{m}\right)$ is proportional to the surface area to mass ratio of the ball. Since the mass of the ball is equal to its volume $\frac{4}{3}\pi R^3$ times its density $\rho$, we have that $$\left(\frac{\pi R^2}{m}\right)=\frac{3}{4\rho R}$$Since R is smaller for the smaller ball than for the larger ball, the expression is larger for the smaller ball than for the larger ball. That means the the effect of air resistance on the ball acceleration is greater for the smaller ball than the larger ball.

It’s very clear now, Thank you!