Question

A ball is thrown from an initial height of 1 meter with an initial upward velocity of 15m/s. The ball's height h (in meters) after t seconds is given by the following.

h=1+15t-5t^2
Find all values of t for which the ball's height is 6 meters. Round your answer(s) to the nearest hundredth.

Answer 1

`\bf \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{h}=1+15t-5t^2}\implies \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{6}=1+15t-5t^2}\implies 0=-5+15t-5t^2 \\\\\\ ~~~~~~~~~~~~\textit{using the quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{5}t^2\stackrel{\stackrel{b}{\downarrow }}{-15}t\stackrel{\stackrel{c}{\downarrow }}{+5}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}`

`\bf t=\cfrac{-(-15)\pm√((-15)^2-4(5)(5))}{2(5)}\implies t=\cfrac{15\pm√(225-100)}{10} \\\\\\ t=\cfrac{15\pm√(125)}{10}\implies t=\cfrac{15\pm√(5^2 \cdot 5)}{10}\implies t=\cfrac{\stackrel{3}{~~\begin{matrix} 15 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\pm ~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~√(5)}{\underset{2}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}`

`\bf t=\cfrac{3\pm √(5)}{2}\implies t= \begin{cases} (3+ √(5))/(2) \approx 2.618\\\\ (3- √(5))/(2)\approx 0.382 \end{cases}`