The air resistance of a sphere is given by²
$$ F_{\textrm{drag}} = \frac{1}{2} \rho C_d A v^2 $$
$C_d$ is usually set to $0.1$ for a sphere¹, $A$ is the relevant surface area, that is, $\left(\frac{\textrm{diametre}}{2}\right)^2 \pi$, $\rho \approx 1.2\textrm{ kgm}^{-3}$³.
We can set up a 1-dimensional coordinate system with "up" being in the positive $s$ direction. Then we have
$$ F_{\textrm{grav}} = - mg $$
and
$$ F_{\textrm{drag}} = - \left(\frac{1}{2} \rho C_d A \equiv K \right) \frac{v^3}{\sqrt{v^2}} $$
as it acts in the direction opposite to $v$. I have introduced a constant $K$ to simplify notation. Then:
$$ -m g - K \frac{v^3}{\sqrt{v^2}} = m \frac{\textrm{d}v}{\textrm{d}t} $$
which is a first-order differential equation in $v$ but second-order in $s$. However, it is not trivial to solve it, as it contains terms non-linear in $v$. Solving the system numerically is also non-trivial, as one of the initial conditions, $v(0)$ is unknown (i.e. we can’t simply evolve it in time from $t = 0$).
The boundary/initial conditions we have are:
$$ s(t = 0) = 0 \quad ; \quad s(t=8.2) = 0 ; \quad v(t=0) = v_0 $$
I would probably go about setting $v(0)$ to some number, then let the system evolve and check where the object is at $t = 8.2$ - if $s$ is positive, decrease $v(0)$, if $s$ is negative, increase $v(0)$ (basically solve the problem numerically).
I think solving the problem numerically would be easier if we had the maximum height $s_{\textrm{max}}$ in the objects path. Setting the time at which it reaches this height to $0$, we would have:
$$ s(0) = s_{\textrm{max}} \quad ; \quad v(0) = 0 $$
and one could simply evolve the system backwards in time until $s(t) = 0$.
I am sorry I cannot give a complete answer, but maybe someone else has an idea on how to continue from here?
Update
John Rennie posted a helpful link which claims to have an analytic solution to this problem. I did not verify said solution, but picked out two formulae:
$$ t_{\textrm{imp}} = \tau \cosh^{-1}\left( \exp\left( \frac{y_{\textrm{peak}}}{\tau v_t}\right) \right)$$
$$ y_{\textrm{peak}} = - v_t \tau \ln\left( \cos\left( \tan^{-1}\left( \frac{v_0}{v_t} \right) \right) \right) $$
where $\tau$ is the characterstic time, $v_t$ is the terminal velocity (the maximum velocity a freely-falling object reaches due to air drag opposing gravity) and $t_{\textrm{imp}}$ is the time after which an object reaches the ground again. $v_0$ is the inital velocity we’re looking for.
Rearranging this gives:
$$ \tan \left( \cos^{-1} \left( \exp \left( - \ln \left( \cosh \left( \frac{t_{\textrm{imp}}}{\tau} \right) \right) \right) \right) \right) v_t = v_0 $$
$v_t$ is given as
$$v_t = \sqrt{\frac{2 m g}{C_d \rho A}}$$
and $\tau = v_t / g$. Plugging all this together gives me:
$$ v_0 = 91.032\textrm{ ms}^{-1} = 203\textrm{ mph} $$
For reference, the thing I put into Qalculate is:
tan(acos(e^(−ln(cosh(8.2s × 9.81 N/kg / sqrt( (2× 4ounce × 9.81 N/kg)/(0.1 × 1.29 kg/m^3 × Pi × (1 in)^2)))))))×sqrt( (2× 4ounce × 9.81 N/kg)/(0.1 × 1.29 kg/m^3 × Pi × (1 in)^2))
Assume that a particle is shot upwards with an initial velocity of $v_0$ at $t = 0$. It then experiences a linear drag force of the form $F_\text{drag} = - \alpha v$, where $\alpha$ is a constant of proportionality. The equation of motion is then
$$
F_\text{grav} + F_\text{drag} = ma \quad \Rightarrow \quad - m g - \alpha v = m \frac{dv}{dt} \quad \Rightarrow \quad \frac{dv}{dt} = -g - \frac{\alpha}{m} v.
$$
This is a separable differential equation, and can be re-arranged and integrated:
\begin{align*}
- \frac{dv}{g + \alpha v/m} &= dt \\
- \frac{m}{\alpha} \int_{v_0}^v \frac{dv'}{mg/\alpha + v'} &= \int_{0}^t dt' \\
\ln \left[ v' + \frac{mg}{\alpha} \right]_{v_0}^v &= - \frac{\alpha t}{m} \\
\frac{v + mg/\alpha}{v_0 + mg/\alpha} &= e^{-\alpha t/m}
\end{align*}
$$
\boxed{v(t) = \left( v_0 + \frac{mg}{\alpha} \right) e^{-\alpha t/m} - \frac{mg}{\alpha} }
$$
The solution for $x(t)$ can be then be found by integrating this once more with respect to time. This method can also be used to find the behavior of projectiles under more realistic forms of drag force, such as quadratic drag; the integrals are just a bit harder.
Alternately, if you a relationship between $v$ and $x$, you can use the chain rule to re-write the above equation as
$$
\frac{dv}{dx} \frac{dx}{dt} = - g - \frac{\alpha}{m} v \quad \Rightarrow \quad
\frac{dv}{dx} v = - g - \frac{\alpha}{m} v
$$
Following the same logic, we then get
\begin{align*}
- \frac{m}{\alpha} \int_{v_0}^v \frac{v' dv'}{mg/\alpha + v'} &= \int_{0}^x dx' \\
\left[ v' - \frac{mg}{\alpha} \ln \left( v' + \frac{mg}{\alpha} \right) \right]_{v_0}^v &= - \frac{\alpha x}{m} \\
v - v_0 - \frac{mg}{\alpha} \ln \left( \frac{ v + mg/\alpha}{v_0 + mg/\alpha} \right) &= - \frac{\alpha x}{m}
\end{align*}
This equation can't be solved exactly to obtain $v$ as a function of $x$ (because of the presence of both logarithmic and polynomial terms), but you could figure out the speed at any height using numerical root-finding techniques; you could also solve for the maximum height by setting $v = 0$ and finding the corresponding $x$. And, as before, this technique could be adapted to other forms of the drag force such as quadratic drag.
Best Answer
The ball will always return to the ground at a lower speed than it was launched. This is simply because it must lose energy to air resistance during its flight.
The ball starts off with a kinetic energy $E_0$. Some of this energy is "spent" in lifting the ball (i.e. it is converted into potential energy) and some of it is spent on moving the air out of the way (i.e. air resistance - it is converted to kinetic energy in the air molecules it pushes away). At the top, the ball stops and now has $E_p < E_0$.
It starts back down again, converting $E_p$ back into kinetic energy, but also losing some more energy as it pushes the air away again. It finally arrives back where it started with $E_f < E_p < E_0$. So it is moving slower than when it left.
Note also, that if it is launched high enough, the air resistance on the way down (which increases as its speed increases) will eventually balance the acceleration due to gravity and it will reach terminal velocity. So not only is the final speed at the bottom always lower than the launch speed, it tends towards a maximum value. Beyond a certain launch speed, it always lands at the same speed.