Given
[tex]-16t^2+v_0t+h_0[/tex]
initial velocity = 60 feet per second
initial height = 95 feet
Find
Maximum height attained by the ball
Explanation
we have given
[tex]\begin{gathered} h(t)=-16t^2+60t+95 \\ h^{\prime}(t)=-32t+60 \end{gathered}[/tex]
put h'(t) = 0
[tex]\begin{gathered} -32t+60=0 \\ -32t=-60 \\ t=\frac{60}{32}=1.875sec \end{gathered}[/tex]
to find the maximum height find the value of h(1.875)
[tex]\begin{gathered} h(1.875)=-16(1.875)^2+60(1.875)+95 \\ h(1.875)=-56.25+112.5+95 \\ h(1.875)=-56.25+207.5 \\ h(1.875)=151.25 \end{gathered}[/tex]
Final Answer
Therefore , the maximum height attained by the ball is 151.25 feet
