Question

6 jordon stands on a bridge and throws a stone up in the air and lands in the water. the stones height (in meters above the ground) x seconds after jordon threw it, is modeled by
\\h(x) = -(x - 2)^2 + 144\\

what is the height of the stone at the time it is thrown?

what is the time that the stone hits the water?

Explanation:

⚡ Using what you learned: evaluating functions · factoring and solving quadratic equations

Step 1: Find the initial height

The time the stone is thrown corresponds to \( x = 0 \). Evaluate \( h(0) \):

$$ h(0) = -(0 - 2)^2 + 144 $$

$$ h(0) = -(-2)^2 + 144 $$

$$ h(0) = -4 + 144 = 140 $$

Step 2: Find the time when the stone hits the water

The stone hits the water when its height is \( 0 \) meters, so set \( h(x) = 0 \):

$$ -(x - 2)^2 + 144 = 0 $$

$$ (x - 2)^2 = 144 $$

$$ x - 2 = \pm 12 $$

Solve for \( x \):

$$ x = 2 + 12 = 14 $$

$$ x = 2 - 12 = -10 $$

Since time must be non-negative, we choose \( x = 14 \).

Answer:

• What is the height of the stone at the time it is thrown? 140 meters
• What is the time that the stone hits the water? 14 seconds