An object is launched directly in the air at a speed of 64 feet per second from a platform located 16 feet in the air. The motion of the object can be modeled using the function f(t)=−16t2+64t+16, where t is the time in seconds and f(t) is the height of the object. When, in seconds, will the object reach its maximum height? Do not include units in your answer.

Answer:After 2 seconds the object reach its maximum height of 80 feet.Step-by-step explanation:Consider the provided function.[tex]f(t)=-16t^2+64t+16[/tex]The function is a downward parabola.The object will reach its max height at the vertex of the parabola. The vertex of the parabola is given by [tex](\frac{-b}{2a}, f(\frac{-b}{2a}))[/tex], Where the standard form is [tex]f=at^2+bt+c[/tex]. By comparing the provided function with the standard form.a=-16, b=64 and c=16Thus, the vertex are:[tex]t=\frac{-b}{2a}[/tex][tex]t=\frac{-64}{2(-16)}=\frac{64}{32}[/tex][tex]t=2[/tex]Now substitute the value of t in the provided function.[tex]f(t)=-16(2)^2+64(2)+16[/tex][tex]f(t)=-16(4)+128+16[/tex][tex]f(t)=-64+144[/tex][tex]f(t)=80[/tex]Hence, after 2 seconds the object reach its maximum height of 80 feet.