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A projectile is modeled by: h = -2t2 + 12t + 5

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Answer

a

23 m23 \text{ m}

Step-by-step solution

5 steps

Here are the solutions to the problems:

  1. A projectile is modeled by: h=−2t2+12t+5h = -2t^2 + 12t + 5

Part a· 2 steps

Find the maximum height.

  1. The maximum height occurs at the vertex of the parabola.

    The time tt at the vertex is given by the formula t=−b2at = -\frac{b}{2a}.
    For h=−2t2+12t+5h = -2t^2 + 12t + 5, we have a=−2a = -2 and b=12b = 12. t=−122(−2)=−12−4=3 seconds t = -\frac{12}{2(-2)} = -\frac{12}{-4} = 3 \text{ seconds}

  2. Substitute t=3t=3 into the height equation to find the maximum height.

    $$
h(3) = -2(3)^2 + 12(3) + 5
$$
$$
h(3) = -2(9) + 36 + 5
$$
$$
h(3) = -18 + 36 + 5
$$
$$
h(3) = 18 + 5 = 23 \text{ meters}
$$
The maximum height is:
$$
\boxed{23 \text{ m}}
$$

Part b· 3 steps

Determine the time when the projectile hits the ground.

  1. The projectile hits the ground when its height hh is 0.

    Set the equation to 0. −2t2+12t+5=0 -2t^2 + 12t + 5 = 0

  2. Use the quadratic formula t=−b±b2−4ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

    Here a=−2a = -2, b=12b = 12, c=5c = 5. t=−12±122−4(−2)(5)2(−2) t = \frac{-12 \pm \sqrt{12^2 - 4(-2)(5)}}{2(-2)} t=−12±144+40−4 t = \frac{-12 \pm \sqrt{144 + 40}}{-4} t=−12±184−4 t = \frac{-12 \pm \sqrt{184}}{-4}

  3. Simplify 184\sqrt{184}.

    $$
\sqrt{184} = \sqrt{4 \times 46} = 2\sqrt{46}
$$
$$
t = \frac{-12 \pm 2\sqrt{46}}{-4}
$$
$$
t = \frac{2(-6 \pm \sqrt{46})}{-4}
$$
$$
t = \frac{-6 \pm \sqrt{46}}{-2}
$$
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