A ball is thrown vertically downwards from the top of a cliff, and its position is tracked from…

Given $S(t) = at^2 + bt + c$. Using the observations: - At $t = 1$: $S(1) = a(1)^2 + b(1) + c = 285 \Rightarrow a + b + c = 285$ - At $t = 2$: $S(2) = a(2)^2 + b(2) + c = 260 \Rightarrow 4a + 2b + c = 260$ - At $t = 4$: $S(4) = a(4)^2 + b(4) + c = 180 \Rightarrow 16a + 4b + c = 180$ This system can be written in matrix form as $AX = B$: $\begin{bmatrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 16 & 4 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 285 \\ 260 \\ 180 \end{bmatrix}$. Determinant of $A$: $|A| = 1(2 - 4) - 1(4 - 16) + 1(16 - 32) = -2 + 12 - 16 = -6 \neq 0$. Since $|A| \neq 0$, $A^{-1}$ exists, where $A^{-1} = \frac{1}{|A|}\text{adj}(A)$. Cofactors of $A$: $C_{11} = -2$, $C_{12} = 12$, $C_{13} = -16$ $C_{21} = 3$, $C_{22} = -15$, $C_{23} = 12$ $C_{31} = -1$, $C_{32} = 3$, $C_{33} = -2$ $\text{adj}(A) = \begin{bmatrix} -2 & 3 & -1 \\ 12 & -15 & 3 \\ -16 & 12 & -2 \end{bmatrix}$. $A^{-1} = -\frac{1}{6} \begin{bmatrix} -2 & 3 & -1 \\ 12 & -15 & 3 \\ -16 & 12 & -2 \end{bmatrix}$. Now $X = A^{-1}B$: $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = -\frac{1}{6} \begin{bmatrix} -2 & 3 & -1 \\ 12 & -15 & 3 \\ -16 & 12 & -2 \end{bmatrix} \begin{bmatrix} 285 \\ 260 \\ 180 \end{bmatrix}$ $= -\frac{1}{6} \begin{bmatrix} -2(285) + 3(260) - 1(180) \\ 12(285) - 15(260) + 3(180) \\ -16(285) + 12(260) - 2(180) \end{bmatrix}$ $= -\frac{1}{6} \begin{bmatrix} -570 + 780 - 180 \\ 3420 - 3900 + 540 \\ -4560 + 3120 - 360 \end{bmatrix} = -\frac{1}{6} \begin{bmatrix} 30 \\ 60 \\ -1800 \end{bmatrix} = \begin{bmatrix} -5 \\ -10 \\ 300 \end{bmatrix}$. Thus, $a = -5, b = -10, c = 300$. The height equation is $S(t) = -5t^2 - 10t + 300$. The height of the cliff corresponds to the initial position at $t = 0$: $S(0) = 300\text{ m}$. Therefore, the height of the cliff is 300 m.