The equation thatmodels the movement of the objet is:
[tex]h(t)=16t^2+60t[/tex]
where h(t) is equal to the hight and t is the time, so we can replace h(t)=36 so:
[tex]\begin{gathered} 36=16t^2+60t \\ -16t^2+60t-36=0 \end{gathered}[/tex]
now we can use the cuadratic function to find the values of t so:
[tex]\begin{gathered} \frac{-b\pm\sqrt[]{b^2-4(a)(c)}}{2a} \\ a=-16 \\ b=60 \\ c=-36 \end{gathered}[/tex]
So the equation become:
[tex]\frac{-60\pm\sqrt[]{60^2-4(-16)(-36)}}{2(-16)}[/tex]
Now we operate so:
[tex]\begin{gathered} \frac{-60\pm\sqrt[]{3600-2304}}{-32} \\ \frac{-60\pm\sqrt[]{1296}}{-32} \\ \frac{-60\pm36}{-32} \end{gathered}[/tex]
Now we solve bout solutions:
[tex]\begin{gathered} t_1=\frac{-60+36}{-32}=0.75 \\ t_2=\frac{-60-36}{-32}=3 \end{gathered}[/tex]
So he will reach a hight of 36 feets at 0.75 seconds and a time of 3 seconds because the first time is when his is going up and the second time is we he is falling back to the grownd.
