A model rocket is launched with an initial upward velocity of 113 ft/s. The rocket’s height h (in feet) after t seconds is giving by the following. h=113t-16t^2 Find all values of t for which the rockets height is 39 feet. Round your answer(s) to the nearest hundredth.

Respuesta :

[tex]\begin{gathered} h=113t-16t^2 \\ h=39ft \\ 39=113t-16t^2 \\ 16t^2-113t+39=0 \\ Use\text{ quadratic formula to find t} \\ \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ a=16 \\ b=-113 \\ c=39 \\ \frac{-(-113)\pm\sqrt{-113^2-4\times16\times39}}{2\times16} \\ \frac{113\pm\sqrt{12769-2496}}{32} \\ \frac{113\pm\sqrt{10273}}{32} \\ \frac{113\pm101.355808911}{32} \\ \frac{113+101.355808911}{32}0r\text{ }\frac{113-101.355808911}{32} \\ \frac{214.355808911}{32}\text{ or }\frac{11.644191089}{32} \\ 0.36388097153\text{ or }6.69861902847 \\ t\approx0.36\text{ or 6.70} \end{gathered}[/tex]