The acceleration due to gravity is g = -9.8 m/s².

Using the first equation of motion, we have:

$$v = u + at$$

>>where:

u is the initial velocity (12 m/s)

v is the final velocity (0 m/s at the maximum height)

a is the acceleration due to gravity (-9.8 m/s²)

t is the time taken (we want to find this)

Substituting the values, we get:

$$0 = 12 \text{ m/s} + (-9.8 \text{ m/s}^2) t$$

Solving for t, we get:

$$t = \frac{12 \text{ m/s}}{9.8 \text{ m/s}^2} \approx 1.22 \text{ s}$$

(b) Maximum height reached:

At the maximum height, the velocity of the ball becomes 0 m/s. Using the second equation of motion, we have:

$$s = ut + \frac{1}{2}at^2$$

where:

s is the maximum height reached

u is the initial velocity (12 m/s)

a is the acceleration due to gravity (-9.8 m/s²)

t is the time taken to reach maximum height (1.22 s)

Substituting the values, we get:

$$s = (12 \text{ m/s})(1.22 \text{ s}) + \frac{1}{2}(-9.8 \text{ m/s}^2)(1.22 \text{ s})^2$$

$$s \approx 7.45 \text{ m}$$

Therefore, the maximum height reached by the ball is approximately 7.45 meters.