The height of the ball, h in meters, is modeled by the next equation:
[tex]h(t)=-t^2+6t+16[/tex]where t is time in seconds.
a. At t = 0, the height of the ball is equal to the height of the cliff. Substituting t = 0 into the equation, we get:
[tex]\begin{gathered} h(0)=-0^2+6\cdot0+16 \\ h(0)=16\text{ m} \end{gathered}[/tex]The height of the cliff is 16 m.
b. When the ball hits the ground, h(t) = 0. Then, we need to find the zeros of the polynomial. Using the quadratic formula with a = -1, b = 6, and c = 16, we get:
[tex]\begin{gathered} t_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t_{1,2}=\frac{-6\pm\sqrt[]{6^2-4\cdot(-1)\cdot16}}{2\cdot(-1)} \\ t_{1,2}=\frac{-6\pm\sqrt[]{100}}{-2} \\ t_1=\frac{-6+10}{-2}=-2 \\ t_2=\frac{-6-10}{-2}=8 \end{gathered}[/tex]In the context of the problem, a negative value has no sense, then t = -2 is discarded.
The ball hits the ground after 8 seconds.
c. The maximum height of the ball corresponds to the vertex of the parabola. The x-coordinate of the vertex, which in this case corresponds to variable time, is computed as follows:
[tex]t_{\max }=\frac{-b}{2a}[/tex]Substituting with a = -1, and b = 6, we get:
[tex]\begin{gathered} t_{\max }=\frac{-6}{2\cdot(-1)} \\ t_{\max }=3\text{ seconds} \end{gathered}[/tex]The height of the ball at t = 3 is:
[tex]\begin{gathered} h(3)=-3^2+6\cdot3+16 \\ h(3)=-9+18+16 \\ h(3)=25\text{ m} \end{gathered}[/tex]The maximum height of the ball is 25 m. This height is reached after 3 seconds.