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If force equals mass times acceleration, wouldn't a basketball dropped from the top of the Eiffel tower exert the same force on the ground as a basketball dropped a foot off the ground? They both have the same mass, and they both are accelerating towards the ground at a rate of $g = 9.81\,{\rm m/s^2}$. (I don't know what terminal velocity is that well as I'm only in physics 1 in high school, but just assume that air drag is not important and the ball doesn't reach terminal velocity.)
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Also, if a ball is dropped high enough to reach terminal velocity, then it accelerates at $0\,{\rm m/s^2}$, so it has a force of ZERO when it hits the ground?
[Physics] Physics Paradox about Newtons Second Law $F=ma$
collisiondragforcesnewtonian-gravitynewtonian-mechanics
Related Solutions
Yes there will be a drag torque opposite the direction of spin. The name for this seems to be viscous torque. According to this paper, the viscous torque on a spinning sphere of radius $R$ in a fluid with viscosity $\eta$ spinning with constant angular velocity $\vec\Omega$ is $$ \vec\tau = -8\pi R^3\eta\vec\Omega $$ The paper goes on to describe how to generalize this relationship to the case of a sphere rotating with arbitrary time-varying angular velocities (under some other assumptions).
If we, however, assume that to good approximation, the viscous torque is linear in the instantaneous angular velocity, then the sphere under the influence of a constant external torque $\vec\tau_0$ will, as you point out, reach terminal angular velocity when the external torque and viscous torque are equal in magnitude but opposite in direction $$ \vec\tau=-\vec\tau_0 $$ This follows from the fact that the net torque $\vec\tau_{\mathrm{net}}$ on the sphere (undergoing fixed axis rotation) is related to its angular acceleration $\vec\alpha$ and moment of inertia $I$ about it's rotation axis by $$ \vec\tau_\mathrm{net} = I\vec\alpha $$ so if the net external torque is zero (which happens when the viscous torque equals the other external torque), then $$ \vec\alpha = 0 $$ the angular acceleration vanishes, so the sphere will no longer speed up in its spinning.
Yes, although I don't think it's totally obvious that your statements are true. Let's assume that the drag force on a given object is \begin{align} \mathbf F_\mathrm{drag} = -\frac{1}{2}\rho AC_dv\mathbf v \end{align} where $\rho$ is the mass density of the fluid in which it moves, $A$ is its cross-sectional area, $C_d$ is its drag coefficient, $\mathbf v$ is its velocity, and $v=|\mathbf v|$ is its speed. Then Newton's Second Law gives the following equation of motion for an object falling near the surface of the Earth under the influence of gravity: \begin{align} ma = m g -\frac{1}{2}\rho AC_d v^2 \end{align} So that the acceleration of the object is \begin{align} a = g - \frac{1}{2}\frac{\rho AC_d}{m}v^2 \end{align} In particular, for a fixed cross-sectional area, increasing the mass of the object will increase its acceleration because the second term will be smaller in magnitude. But that also means that the object's speed with increase faster, so that the second term will grow faster; there are competing affects. So which one wins out? Well, the equation of motion can be looked upon as a differential equation for the velocity $v(t)$ as a function of time; \begin{align} \dot v(t) = g-\frac{1}{2}\frac{\rho AC_d}{m} v(t)^2 \end{align} With the initial condition $v(0) = 0$, namely if you just drop the object, the solution (thanks to to Stephen Wolfram) is \begin{align} v(t) = \sqrt{\frac{2gm}{AC_d\rho}}\tanh\left(\sqrt\frac{AC_dg\rho}{2m}t\right) \end{align} Let's plot this function for some different mass values but keeping all other parameters the same
The lowest blue curve corresponds to the lowest mass, and each successive curve above it corresponds to a mass twice as large as for the last curve.
It's clear that the terminal velocities of the more massive objects are greater and that these velocities are achieved at a later time. Moreover, after terminal velocity is reached, the object no longer accelerates.
Best Answer
I think the other two answers may have overlooked the source of your confusion, which is quite simple.
The $F$ in $F=ma$ is the force being exerted on the object of mass $m$ to give it the acceleration $a$, not the force that that object will exert when it hits something.
In the case of your example, the force of gravity on the basketball is independent of the height. The force that the basketball exerts on the ground is an entirely different matter.