1. **State the problem:** We have a table showing the height $y$ of a baseball at different times $x$ seconds after it was hit. We want to predict the height after 5 seconds using the given data. 2. **Analyze the data:** The points are $(0,3)$, $(0.5,39)$, $(1,67)$, $(1.5,87)$, and $(2,99)$. The height increases quickly but the rate of increase slows down, suggesting a quadratic or similar curve. 3. **Choose a model:** We use a quadratic function $y = ax^2 + bx + c$ because projectile motion typically follows a parabola. 4. **Set up equations using known points:** - At $x=0$, $y=3$ gives $c=3$. - At $x=0.5$, $y=39$ gives $a(0.5)^2 + b(0.5) + 3 = 39$. - At $x=1$, $y=67$ gives $a(1)^2 + b(1) + 3 = 67$. 5. **Write the system:** $$ \begin{cases} 0.25a + 0.5b + 3 = 39 \\ a + b + 3 = 67 \end{cases} $$ 6. **Simplify:** $$ \begin{cases} 0.25a + 0.5b = 36 \\ a + b = 64 \end{cases} $$ 7. **Solve the system:** Multiply the first equation by 4: $$ \cancel{4}(0.25a + 0.5b) = \cancel{4} \times 36 \Rightarrow a + 2b = 144 $$ Subtract the second equation $a + b = 64$ from this: $$ (a + 2b) - (a + b) = 144 - 64 \Rightarrow b = 80 $$ Use $a + b = 64$: $$ a + 80 = 64 \Rightarrow a = -16 $$ 8. **Write the quadratic model:** $$ y = -16x^2 + 80x + 3 $$ 9. **Predict height at $x=5$ seconds:** $$ y = -16(5)^2 + 80(5) + 3 = -16(25) + 400 + 3 = -400 + 400 + 3 = 3 $$ 10. **Interpretation:** The predicted height after 5 seconds is 3 feet. 11. **Answer to part b:** The actual height after 5 seconds is about 3 feet, matching our prediction. Differences in real life could be due to air resistance, spin, or measurement errors, but here the model fits well. **Final answer:** After 5 seconds, the height will be about $3$ feet.