Question

A ball is thrown from an initial height of 3 meters with an initial upward velocity of 30 m/s. The ball's height h (in meters) after seconds is given by the
following
h = 3+302-57
Find all values of tfor which the ball's height is 13 meters.
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
1 seconds
х
5
?
Initial
height
ground

Answer 1

$h=3+30t-5t^2\implies 13=3+30t-5t^2\implies 0=-10+30t-5t^2 \\\\\\ 5t^2-30t+10=0\implies 5(t^2-6t+2)=0\implies t^2-6t+2=0$

now, it doesn't give us nice looking integers, so let's plug that into the quadratic formula.

$~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{1}t^2\stackrel{\stackrel{b}{\downarrow }}{-6}t\stackrel{\stackrel{c}{\downarrow }}{+2}=0 ~\hspace{10em} t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t= \cfrac{ - (-6) \pm \sqrt { (-6)^2 -4(1)(2)}}{2(1)}\implies t=\cfrac{6\pm\sqrt{36-8}}{2}$

$t=\cfrac{6\pm\sqrt{28}}{2}\implies t=\cfrac{6\pm\sqrt{2^2\cdot 7}}{2}\implies t=\cfrac{6\pm 2\sqrt{7}}{2}\implies t=\cfrac{2(3\pm 1\sqrt{7})}{2} \\\\\\ t=3\pm \sqrt{7}\implies t= \begin{cases} 3+\sqrt{7}\\ 3-\sqrt{7} \end{cases}\implies t\approx \begin{cases} 5.65\\ 0.35 \end{cases}$