Given
[tex]h\mleft(t\mright)=44t-16t^{2.}[/tex]
Find
the maximum height that the ball will reach
Explanation
The maximum of the function can be found by equating its first derivative to zero.
so, we need to find the first derivative of the function.
[tex]\begin{gathered} h\mleft(t\mright)=44t-16t^{2.} \\ h^{\prime}(t)=44-32t \end{gathered}[/tex]
put h'(t) = 0
[tex]\begin{gathered} 44-32t=0 \\ 32t=44 \\ t=\frac{44}{32} \\ \\ t=\frac{11}{8} \end{gathered}[/tex]
so , the maximum height is found by substituting the value of t into the original equation
[tex]\begin{gathered} h(t)=44(\frac{11}{8})-16(\frac{11}{8})^2 \\ \\ h(t)=60.5+30.25 \\ h(t)=30.25 \end{gathered}[/tex]
Final Answer
the maximum height that the ball will reach is 30.25 feet
