$$v^2 = u^2 + 2gs$$
where:
- v is the final velocity of the projectile (at the maximum height, it will be 0 m/s)
- u is the initial velocity of the projectile (12 m/s)
- g is the acceleration due to gravity (-10 m/s²)
- s is the displacement of the projectile (in this case, the maximum height, h)
Substituting the given values into the equation:
$$0^2 = (12 \text{ m/s})^2 + 2(-10 \text{ m/s}^2)h$$
Simplifying:
$$0 = 144 \text{ m}^2/\text{s}^2 - 20h \text{ m/s}^2$$
$$20h \text{ m/s}^2 = 144 \text{ m}^2/\text{s}^2$$
Solving for h:
$$h = \frac{144 \text{ m}^2/\text{s}^2}{20 \text{ m/s}^2}$$
$$h = 7.2 \text{ m}$$
Therefore, the maximum height reached by the arrow is 7.2 meters.
