Question

an archer shoots an arrow up into the air. the height h(t) in meters of the arrow after t seconds is modeled by h(t)= - 9.8t² + 32t + 3. graph the function on a graphing calculator and round all answers to the second - decimal place (hundredths). find h(2). what does h(2) mean in context of the problem? how long will it take for the arrow to hit the ground after it is fired? at what time(s) will the arrow be 10 meters in the air?

Explanation:

Step1: Calculate \(h(2)\)

Substitute \(t = 2\) into \(h(t)\): \(h(2)=-9.8\times(2)^{2}+32\times2 + 3\).

Step2: Simplify \(h(2)\)

Calculate powers and multiplications: \(-9.8\times4=-39.2\), \(32\times2 = 64\), then \(h(2)=-39.2+64 + 3\).

Step3: Solve for \(t\) when \(h(t)=10\)

Set up quadratic equation \(-9.8t^{2}+32t - 7 = 0\) and use quadratic formula.

Step4: Calculate discriminant for \(h(t)=10\)

\(\Delta=(32)^{2}-4\times(-9.8)\times(-7)=1024 - 274.4 = 749.6\).

Step5: Find \(t\) values for \(h(t)=10\)

Use \(t=\frac{-b\pm\sqrt{\Delta}}{2a}\) to get \(t\approx0.24\) and \(t\approx3.03\) seconds.

Step6: Solve for \(t\) when \(h(t)=0\)

Set up quadratic equation \(-9.8t^{2}+32t + 3 = 0\) and use quadratic formula.

Step7: Calculate discriminant for \(h(t)=0\)

\(\Delta=(32)^{2}-4\times(-9.8)\times3=1024 + 117.6 = 1141.6\).

Step8: Find \(t\) value for \(h(t)=0\)

Use \(t=\frac{-b\pm\sqrt{\Delta}}{2a}\), reject negative value, get \(t\approx3.36\) seconds.

Answer:

1. Find \(h(2)\):

• Substitute \(t = 2\) into the function \(h(t)=-9.8t^{2}+32t + 3\).
• \(h(2)=-9.8\times(2)^{2}+32\times2 + 3\).
• First, calculate \((2)^{2}=4\). Then, \(-9.8\times4=-39.2\), and \(32\times2 = 64\).
• \(h(2)=-39.2+64 + 3\).
• \(h(2)=27.8\) meters.

1. What does \(h(2)\) mean in the context of the problem?:

• \(h(2)\) represents the height of the arrow in meters \(2\) seconds after it is shot into the air.

1. At what time(s) will the arrow be 10 meters in the air?:

• Set \(h(t)=10\), so \(-9.8t^{2}+32t + 3 = 10\).
• Rearrange the equation to get \(-9.8t^{2}+32t - 7=0\).
• Use the quadratic - formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for a quadratic equation \(ax^{2}+bx + c = 0\). Here, \(a=-9.8\), \(b = 32\), and \(c=-7\).
• First, calculate the discriminant \(\Delta=b^{2}-4ac=(32)^{2}-4\times(-9.8)\times(-7)\).
• \((32)^{2}=1024\), and \(4\times9.8\times7 = 274.4\).
• \(\Delta=1024 - 274.4 = 749.6\).
• Then, \(t=\frac{-32\pm\sqrt{749.6}}{2\times(-9.8)}=\frac{-32\pm27.37}{-19.6}\).
• For the plus - sign: \(t=\frac{-32 + 27.37}{-19.6}=\frac{-4.63}{-19.6}\approx0.24\) seconds.
• For the minus - sign: \(t=\frac{-32-27.37}{-19.6}=\frac{-59.37}{-19.6}\approx3.03\) seconds.

1. How long will it take for the arrow to hit the ground after it is fired?:

• Set \(h(t)=0\), so \(-9.8t^{2}+32t + 3 = 0\).
• Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a=-9.8\), \(b = 32\), and \(c = 3\).
• Calculate the discriminant \(\Delta=b^{2}-4ac=(32)^{2}-4\times(-9.8)\times3\).
• \((32)^{2}=1024\), and \(4\times9.8\times3 = 117.6\).
• \(\Delta=1024+117.6 = 1141.6\).
• Then, \(t=\frac{-32\pm\sqrt{1141.6}}{-19.6}=\frac{-32\pm33.79}{-19.6}\).
• For the plus - sign: \(t=\frac{-32 + 33.79}{-19.6}=\frac{1.79}{-19.6}\approx - 0.09\) (rejected since time cannot be negative).
• For the minus - sign: \(t=\frac{-32-33.79}{-19.6}=\frac{-65.79}{-19.6}\approx3.36\) seconds.