The given expression in the question is
[tex]h=6+21t-16t^2[/tex]
with the value of h given as
[tex]h=12ft[/tex]
By equating both equations, we will have
[tex]\begin{gathered} 12=6+21t-16t^2 \\ 12-6-21t+16t^2=0 \\ 6-21t+16t^2=0 \\ 16t^2-21t+6=0 \end{gathered}[/tex]
To find the value of t we will use the quadratic formula of
[tex]ax^2+bx+c=0[/tex]
which is
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{where} \\ a=16 \\ b=-21 \\ c=6 \end{gathered}[/tex]
By substitution, we will have
[tex]\begin{gathered} t=\frac{-(-21)\pm\sqrt[]{(-21)^2-4\times16\times6}}{2\times16} \\ t=\frac{21\pm\sqrt[]{441-384}}{32} \\ t=\frac{21\pm\sqrt[]{57}}{32} \\ t=\frac{21\pm7.5498}{32} \\ t=\frac{21+7.5498}{32}\text{ or t=}\frac{21-7.5498}{32} \\ t=\frac{28.5498}{32\text{ }}\text{ or }t=\frac{13.4502}{32\text{ }} \\ t=0.89\text{ or t=0.42} \end{gathered}[/tex]
Alternatively, Solving the equation graphically we will have
Therefore,
The values of t( to the nearest hundredth) t= 0.89sec or 0.42sec
