Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.6 Parametric Equations, Graphs, and Applications - 8.6 Exercises - Page 399: 42b

$y = 0.554~x-0.00113~x^2+2.5$

From Part (a): $x = 118.95~t$ $y = 65.93~t-16~t^2+2.5$ We can find an expression for $t$ in terms of $x$: $x = 118.95~t$ $t = \frac{x}{118.95}$ We can substitute this value in the equation for $y$: $y = 65.93~t-16~t^2+2.5$ $y = 65.93~(\frac{x}{118.95})-16~(\frac{x}{118.95})^2+2.5$ $y = 0.554~x-0.00113~x^2+2.5$