Respuesta :
The equation that defines the motion is given as:
[tex]h=-16t^2+110t+20[/tex]To find when the height will be 59 feet, substitute h=59 into the equation and solve for t:
[tex]\begin{gathered} 59=-16t^2+110t+20 \\ \text{Swap the sides of the equation:} \\ -16t^2+110t+20=59 \\ -16t^2+110t+20-59=0 \\ -16t^2+110t-39=0 \\ \text{Multiply both sides of the equation by -1:} \\ 16t^2-110t+39=0 \end{gathered}[/tex]Solve the quadratic equation using any method of your choice.
Using the quadratic formula where a=16, b=-110 and c=39:
[tex]\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t=\frac{-(-110)\pm\sqrt[]{(-110)^2-4(16)(39)}}{2(16)} \\ t=\frac{110\pm\sqrt[]{12100-2496}}{32} \\ t=\frac{110\pm\sqrt[]{9604}}{32}_{} \\ t=\frac{110\pm98}{32} \\ t=\frac{208}{32}or\text{ }\frac{12}{32} \\ t=6.5s\text{ or }0.375s \end{gathered}[/tex]The unit of time is seconds.
T
The object will reach the ground when the height is zero, that is when h=0.
Substitute into the equation and solve for t again:
[tex]0=-16t^2+110t+20[/tex]Using the same procedure, where a=-16, b=110 and c=20, use the quadratic formula to solve for t to get:
[tex]t\approx-0.18s\text{ or }t\approx7.05s[/tex]Since time cannot be negative, discard the negative value. So, the object will reach the ground at about t=7.05 seconds.
