Answer:The maximum height of the baseball is 9 feetStep-by-step explanation:we have[tex]p(t)=\frac{1}{2}gt^2+v_0t+p_0[/tex]wherep(t) ----> baseball position above the ground in feett ----> the time in secondsv_0 ----> is the initial velocity in ft/secp_0 ---> initial position above the ground we have[tex]g=-32\frac{ft}{sec^2} \\\\v_0=16\frac{ft}{sec}\\\\p_0=5\ ft[/tex]substitute the given values[tex]p(t)=\frac{1}{2}(-32)t^2+16t+5[/tex][tex]p(t)=-16t^2+16t+5[/tex]This is the equation of a vertical parabola open downwardThe vertex represent a maximumConvert the quadratic equation in vertex form[tex]p(t)=-16t^2+16t+5[/tex]Factor -16 leading coefficient[tex]p(t)=-16(t^2-t)+5[/tex]Complete the square[tex]p(t)=-16(t^2-t+\frac{1}{4})+5+4[/tex][tex]p(t)=-16(t^2-t+\frac{1}{4})+9[/tex]Rewrite as perfect squares[tex]p(t)=-16(t-\frac{1}{2})^2+9[/tex]The vertex is the point (0.5,9)The maximum height of the baseball above the ground is the y-coordinate of the vertexthereforeThe maximum height of the baseball is 9 feet