Initial height, s0 = 0.5 m

The acceleration is given by $a = \frac{dv}{dt}$.

$$v = \int a \, dt$$ $$v = \int -10 \, dt$$ $$v = -10t + C$$

At $t=0$, the initial velocity is $v = 16$ m/s.

$$16 = -10(0) + C$$ $$C = 16$$

So, the expression for velocity is:

$$\boxed{v = 16 - 10t \, \text{m/s}}$$

The velocity is given by $v = \frac{ds}{dt}$.

$$s = \int v \, dt$$ $$s = \int (16 - 10t) \, dt$$ $$s = 16t - \frac{10t^2}{2} + D$$ $$s = 16t - 5t^2 + D$$

At $t=0$, the initial height is $s = 0.5$ m.

$$0.5 = 16(0) - 5(0)^2 + D$$ $$D = 0.5$$

So, the expression for height above the ground is:

$$\boxed{s = 0.5 + 16t - 5t^2 \, \text{m}}$$

Maximum height occurs when the velocity of the ball is momentarily zero ($v=0$).
Set the velocity expression from part (a) to zero:

$$16 - 10t = 0$$ $$10t = 16$$ $$t = \frac{16}{10} = 1.6 \, \text{s}$$

$$s_{max} = 0.5 + 16(1.6) - 5(1.6)^2$$ $$s_{max} = 0.5 + 25.6 - 5(2.56)$$ $$s_{max} = 0.5 + 25.6 - 12.8$$ $$s_{max} = 26.1 - 12.8$$ $$\boxed{s_{max} = 13.3 \, \text{m}}$$