Write out the function given in the question
[tex]h(t)=56t-16t^2[/tex]
Note that at the maximum height the derivative of the function would be equal to zero
Find the derivative of the function using the formula
[tex]\begin{gathered} \frac{dh}{\text{ dt}}=ant^{n-1} \\ \frac{dh}{\text{ dt}}=56\times1t^{1-1}-16\times2t^{2-1} \\ =56-32t \end{gathered}[/tex]
Equate the derivative to zero to get the value of t because the derivative is zero at maximum height
[tex]\begin{gathered} 56-32t=0 \\ 56=32t \\ 32t=56 \\ t=\frac{56}{32} \\ t=\frac{7}{4} \end{gathered}[/tex]
Substitute the value of t into the function
[tex]\begin{gathered} h(t)=56t-16t^2 \\ \text{when t=7/4} \\ h(\frac{7}{4})=56(\frac{7}{4})-16(\frac{7}{4})^2 \end{gathered}[/tex][tex]\begin{gathered} h(\frac{7}{4})=14\times7-16(\frac{49}{16}) \\ =98-49=49\text{ feet} \end{gathered}[/tex]
Hence, the maximum height that the ball will reach is 49feet
