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A ball travels on a parabolic path in which the height (in feet) is given by the expression $-16t^2+80t+21$, where $t$ is the time after launch. What is the maximum height of the ball?

Guest Jan 29, 2018
edited by Guest  Jan 29, 2018

#1
+19853
+2

A ball travels on a parabolic path in which the height (in feet) is given by the expression $-16t^2+80t+21$,
where $t$ is the time after launch.
What is the maximum height of the ball?

The graph:

$$\begin{array}{|lrcll|} \hline & h(t) &=& -16t^2+80t+21 \\ h = 0 \ ? \\ & -16t^2+80t+21 &=& 0 \\\\ & t_{1,2} &=& \dfrac{-80 \pm\sqrt{80^2-4\cdot(-16)\cdot21}}{2\cdot(-16)} \\\\ & &=& \dfrac{-80 \pm\sqrt{6400+1344} }{-32} \\\\ & t_{1,2} &=& \dfrac{-80 \pm88}{-32} \\\\ & t_1 &=& \dfrac{-80 + 88}{-32} \\\\ & &=& \dfrac{8}{-32} \\\\ & \mathbf{t_1} &\mathbf{=}& \mathbf{-0.25} \\\\ & t_2 &=& \dfrac{-80 - 88}{-32} \\\\ & &=& \dfrac{168}{32} \\\\ & \mathbf{t_2} &\mathbf{=}& \mathbf{5.25} \\\\ &t_{h_\text{max}} &=& \dfrac{t_1+t_2}{2} \\\\ & &=& \dfrac{-0.25+5.25}{2} \\\\ & &=& \dfrac{5}{2} \\\\ &\mathbf{t_{h_\text{max}}} &\mathbf{=}& \mathbf{2.5} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline h_\text{max} &=& -16\mathbf{t_{h_\text{max}}}^2+80\mathbf{t_{h_\text{max}}}+21 \quad & | \quad \mathbf{t_{h_\text{max}} = 2.5} \\ &=& -16\cdot 2.5^2+80\cdot 2.5+21 \\ &=& -100+200+21 \\ &\mathbf{=}& \mathbf{121~ \text{feet}}\\ \hline \end{array}$$

The maximum height of the ball is $$\mathbf{121~ \text{feet}}$$

heureka  Jan 29, 2018
#1
+19853
+2

A ball travels on a parabolic path in which the height (in feet) is given by the expression $-16t^2+80t+21$,
where $t$ is the time after launch.
What is the maximum height of the ball?

The graph:

$$\begin{array}{|lrcll|} \hline & h(t) &=& -16t^2+80t+21 \\ h = 0 \ ? \\ & -16t^2+80t+21 &=& 0 \\\\ & t_{1,2} &=& \dfrac{-80 \pm\sqrt{80^2-4\cdot(-16)\cdot21}}{2\cdot(-16)} \\\\ & &=& \dfrac{-80 \pm\sqrt{6400+1344} }{-32} \\\\ & t_{1,2} &=& \dfrac{-80 \pm88}{-32} \\\\ & t_1 &=& \dfrac{-80 + 88}{-32} \\\\ & &=& \dfrac{8}{-32} \\\\ & \mathbf{t_1} &\mathbf{=}& \mathbf{-0.25} \\\\ & t_2 &=& \dfrac{-80 - 88}{-32} \\\\ & &=& \dfrac{168}{32} \\\\ & \mathbf{t_2} &\mathbf{=}& \mathbf{5.25} \\\\ &t_{h_\text{max}} &=& \dfrac{t_1+t_2}{2} \\\\ & &=& \dfrac{-0.25+5.25}{2} \\\\ & &=& \dfrac{5}{2} \\\\ &\mathbf{t_{h_\text{max}}} &\mathbf{=}& \mathbf{2.5} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline h_\text{max} &=& -16\mathbf{t_{h_\text{max}}}^2+80\mathbf{t_{h_\text{max}}}+21 \quad & | \quad \mathbf{t_{h_\text{max}} = 2.5} \\ &=& -16\cdot 2.5^2+80\cdot 2.5+21 \\ &=& -100+200+21 \\ &\mathbf{=}& \mathbf{121~ \text{feet}}\\ \hline \end{array}$$

The maximum height of the ball is $$\mathbf{121~ \text{feet}}$$

heureka  Jan 29, 2018