A ball is thrown into the air with an upward velocity of 45 ft/s. Its height (h) in feet after t seconds is given by the function h = -16t^2 + 45t +6. After about how many seconds will the ball hit the ground?
+33661
The ground is at height h = 0, so solve the quadratic equation 0 = -16t2 + 45t +6
$${\mathtt{\,-\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{45}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{2\,409}}}}{\mathtt{\,-\,}}{\mathtt{45}}\right)}{{\mathtt{32}}}}\\
{\mathtt{t}} = {\frac{\left({\sqrt{{\mathtt{2\,409}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{45}}\right)}{{\mathtt{32}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = -{\mathtt{0.127\: \!548\: \!898\: \!976\: \!003\: \!3}}\\
{\mathtt{t}} = {\mathtt{2.940\: \!048\: \!898\: \!976\: \!003\: \!3}}\\
\end{array} \right\}$$
The only positive solution is t ≈ 2.94 seconds.
.
+33661
The ground is at height h = 0, so solve the quadratic equation 0 = -16t2 + 45t +6
$${\mathtt{\,-\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{45}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{2\,409}}}}{\mathtt{\,-\,}}{\mathtt{45}}\right)}{{\mathtt{32}}}}\\
{\mathtt{t}} = {\frac{\left({\sqrt{{\mathtt{2\,409}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{45}}\right)}{{\mathtt{32}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = -{\mathtt{0.127\: \!548\: \!898\: \!976\: \!003\: \!3}}\\
{\mathtt{t}} = {\mathtt{2.940\: \!048\: \!898\: \!976\: \!003\: \!3}}\\
\end{array} \right\}$$
The only positive solution is t ≈ 2.94 seconds.
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