ANSWER
the object will hit the ground at 4.793 seonds
STEP-BY-STEP EXPLANATION
Given the function below
[tex]s(t)=-16t^2\text{ + 60t + 80}[/tex]
To find the value of t in seconds, let s(t) = 0
The new equation becomes
[tex]-16t^2\text{ + 60t + 80 = 0}[/tex]
The quadratic function can be solved using the general quadratic formula
From the equation above,
Let a = -16, b = 60, and c = 80
[tex]\begin{gathered} t\text{ = }\frac{-b\text{ }\pm\sqrt[]{b^2\text{ - 4ac}}}{2a} \\ t\text{ = }\frac{-(60)\pm\sqrt[]{60^2\text{ - 4 (-16)(80)}}}{2\text{ (-16)}} \\ t\text{ = }\frac{-60\pm\sqrt[]{3600\text{ + 5120}}}{-32} \\ t\text{ = }\frac{-60\text{ }\pm\sqrt[]{8720}}{-32} \\ t\text{ = }\frac{-60\pm93.38}{-32} \\ t\text{ = }\frac{-60\text{ + 93.38}}{-32}\text{ or }\frac{-60\text{ - 93.38}}{-32} \\ t\text{ = }\frac{33.38}{-32}\text{ or }\frac{-153.8}{-32} \\ t\text{ = -1.043 seconds or 4.793 secons} \end{gathered}[/tex]
Hence, the object will hit the ground at 4.793 seonds
