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)=βˆ’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.

Accepted Solution

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.