An object is thrown upward from the top of a 144-foot building with an initial velocity of 128 feet per second. The height h of the object after t seconds is given by the quadratic equation h equals negative 16 t squared plus 128 t plus 144. When will the object hit the ground?

[tex]t=\frac{-(-128)\pm \sqrt{(-128)^2-4\times 16\times (-144)}}{2\times 16}\\\\t=\frac{128\pm \sqrt{25600}}{32}\\\\t=\frac{128\pm 160}{32}\\\\t=9s\texttt{ or }t=-1s[/tex]

Negative time is not possible, hence after 9 seconds the object reaches ground.

An object is thrown upward from the top of a 144​-foot building with an initial velocity of 128 feet per second. The height h of the object after t seconds is given by the quadratic equation h equals negative 16 t squared plus 128 t plus 144. When will the object hit the​ ground?

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An object is thrown upward from the top of a 144-foot building with an initial velocity of 128 feet per second. The height h of the object after t seconds is given by the quadratic equation h equals negative 16 t squared plus 128 t plus 144. When will the object hit the ground?