Given data:
The height of the desk above the valley is 68.
if an object is droped from the desk the height after t seconds is ,
[tex]h=-16t^2+68[/tex]put h = 4 in the above equation.
[tex]\begin{gathered} h=-16t^2+68 \\ 4=-16t^2+68 \\ 16t^2=68-4 \\ 16t^2=64 \\ t^2=\frac{64}{16} \\ t^2=4 \\ t=\pm\sqrt[]{4} \\ t=\pm2 \\ t=2 \end{gathered}[/tex]t=2 because time can only be positive.
thus, it will take 2 seconds for the object be 4 ft above the valley floor.
