Given,
The initial velocity of the rocket, u=215 ft/s
The equation which gives the height of the rocket at any given instant of time,
[tex]h=215t-16t^2[/tex]The height of the rocket, h=97 ft.
The given equation is a quadratic equation. On rearranging the above equation and substituting the value of h,
[tex]\begin{gathered} 16t^2-215t+h=0 \\ 16t^2-t+97=0 \end{gathered}[/tex]A quadratic equation is written as,
[tex]ax^2+bx+c=0[/tex]Comparing the two equations,
x=t, a=16, b=-215, c=97.
The solution of a quadratic equation is given by,
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]On substituting the known values in the above equation,
[tex]\begin{gathered} t=\frac{-(-215)\pm\sqrt[]{(-215)^2-4\times16\times97}}{2\times16} \\ =\frac{215\pm\sqrt[]{40017}}{32} \\ =12.97\text{ or}\pm0.47\text{ } \end{gathered}[/tex]Thus, the values of t are 12.97 s or 0.47 s.
That is, the rocket will be at a height of 97 feet at t=12.97 s or t=0.47 s