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# HELP ASAP!!

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A ball travels on a parabolic path in which the height (in feet) is given by the expression -25t^2+ 75t +24, where t is the time after launch. At what time is the height of the ball at its maximum?

Oct 21, 2019

#1
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A ball travels on a parabolic path in which the height (in feet) is given by the expression $$-25t^2+ 75t +24$$,

where t is the time after launch.

At what time is the height of the ball at its maximum?

$$\begin{array}{|rcll|} \hline -25t^2+ 75t +24 &=& 0 \\ t &=& \dfrac{ -75 \pm \sqrt{75^2-4\cdot(-25)\cdot 24} }{2\cdot(-25)} \\ \hline t_1 &=& \dfrac{ -75 + \sqrt{75^2-4\cdot(-25)\cdot 24} }{-50} \\\\ t_2 &=& \dfrac{ -75 - \sqrt{75^2-4\cdot(-25)\cdot 24} }{-50} \\ \hline t_{\text{max}} &=& \dfrac{t_1+t_2}{2} \\\\ t_{\text{max}} &=& \dfrac{\dfrac{ -75 + \sqrt{75^2-4\cdot(-25)\cdot 24} }{-50} +\dfrac{ -75 - \sqrt{75^2-4\cdot(-25)\cdot 24} }{-50} }{2} \\\\ t_{\text{max}} &=& \dfrac{ -75 + \sqrt{75^2-4\cdot(-25)\cdot 24}-75 - \sqrt{75^2-4\cdot(-25)\cdot 24} } {(-50)\cdot 2} \\\\ t_{\text{max}} &=& \dfrac{ -150 } { -100 } \\\\ \mathbf{t_{\text{max}}} &=& \mathbf{1.5} \\ \hline \end{array}$$ Oct 22, 2019