Solution:
Given:
[tex]h=-16t^2+97t+17[/tex]when the height is 71ft,
h = 71
Hence,
[tex]\begin{gathered} 71=-16t^2+97t+17 \\ 16t^2-97t+71-17=0 \\ 16t^2-97t+54=0 \end{gathered}[/tex]Using the quadratic formula;
[tex]\begin{gathered} a=16,b=-97,c=54 \\ t=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ t=\frac{-(-97)\pm\sqrt{(-97)^2-(4\times16\times54)}}{2\times16} \end{gathered}[/tex]Hence,
[tex]\begin{gathered} t=\frac{97\pm77.156}{32} \\ t_1=\frac{97+77.156}{32}=\frac{174.156}{32}=5.442375=5.44s \\ t_2=\frac{97-77.156}{32}=\frac{19.844}{32}=0.620125=0.62s \end{gathered}[/tex]Therefore, the height will be 71 feet at 5.44 seconds or 0.62 seconds.
The object will reach the ground at;
The object will reach the height of 17ft at;
h = 17
[tex]\begin{gathered} 17=-16t^2+97t+17 \\ 16t^2-97t=0 \\ t(16t-97)=0 \\ t=0 \\ 16t-97=0 \\ 16t=97 \\ t=\frac{97}{16} \\ t=6.0625s \end{gathered}[/tex]Hence, the time it will take to reach the ground will be the time it takes to go up and return back to the ground;
[tex]\begin{gathered} t=6.0625\times2 \\ t=12.125s \end{gathered}[/tex]It will take 12.125 seconds for the object to reach the ground.