Determine the initial velocity (u) of the stone.

We use the kinematic equation for displacement: $s = ut + \frac{1}{2}at^2$.
Let the upward direction be positive. The initial position is $85 \text{ m}$ above the ground. The final position is $0 \text{ m}$ (ground level). So, the displacement $s = 0 - 85 = -85 \text{ m}$.
The acceleration due to gravity $a = -g = -9.8 \text{ m/s}^2$.
The total time $t = 5 \text{ s}$.

$$s = ut + \frac{1}{2}at^2$$ $$-85 = u(5) + \frac{1}{2}(-9.8)(5)^2$$ $$-85 = 5u - 4.9(25)$$ $$-85 = 5u - 122.5$$ $$5u = 122.5 - 85$$ $$5u = 37.5$$ $$u = \frac{37.5}{5}$$ $$u = 7.5 \text{ m/s}$$

The stone reaches its maximum height when its vertical velocity becomes $0 \text{ m/s}$.
We use the kinematic equation: $v^2 = u^2 + 2as$.
Here, $v = 0 \text{ m/s}$, $u = 7.5 \text{ m/s}$, and $a = -9.8 \text{ m/s}^2$.
Let $h_{add}$ be the additional height reached above the tower.

$$0^2 = (7.5)^2 + 2(-9.8)h_{add}$$ $$0 = 56.25 - 19.6 h_{add}$$ $$19.6 h_{add} = 56.25$$ $$h_{add} = \frac{56.25}{19.6}$$ $$h_{add} \approx 2.87 \text{ m}$$

The greatest height above the ground is the tower's height plus this additional height.

$$H_{ground} = 85 \text{ m} + 2.87 \text{ m}$$ $$H_{ground} = 87.87 \text{ m}$$

The greatest height above the ground is $\boxed{\text{87.87 m}}$.

We use the kinematic equation: $v = u + at$.
Here, $u = 7.5 \text{ m/s}$, $a = -9.8 \text{ m/s}^2$, and $t = 5 \text{ s}$.

$$v = 7.5 + (-9.8)(5)$$ $$v = 7.5 - 49$$ $$v = -41.5 \text{ m/s}$$

The negative sign indicates the velocity is directed downwards.
The velocity with which it reaches the ground is $\boxed{\text{41.5 m/s downwards}}$.

We use the kinematic equation: $v = u + at$.
At maximum height, the final velocity $v = 0 \text{ m/s}$.
Here, $u = 7.5 \text{ m/s}$ and $a = -9.8 \text{ m/s}^2$.

$$0 = 7.5 + (-9.8)t_{max}$$ $$9.8 t_{max} = 7.5$$ $$t_{max} = \frac{7.5}{9.8}$$ $$t_{max} \approx 0.77 \text{ s}$$

The time taken to reach the maximum height is $\boxed{\text{0.77 s}}$.