Question

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 1

2 seconds

Explanation:

When the object reaches its highest point is the t-value of the turning point. Thus we need to find the where maximum quadratic function

−16t2+64t+16

occurs.

We know that the maximum (or minimum) of a parabola is reached at x=−b2a=−642(−16)=2, showing that the object will reach its highest point at 2 seconds.

Answer 2

After 2 seconds the object reach its maximum height of 80 feet.

Explanation:

Consider the provided function.

`f(t)=-16t^2+64t+16`

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
`((-b)/(2a), f((-b)/(2a)))`,

Where the standard form is
`f=at^2+bt+c`.

By comparing the provided function with the standard form.

a=-16, b=64 and c=16

Thus, the vertex are:

`t=(-b)/(2a)`

`t=(-64)/(2(-16))=(64)/(32)`

`t=2`

Now substitute the value of t in the provided function.

`f(t)=-16(2)^2+64(2)+16`

`f(t)=-16(4)+128+16`

`f(t)=-64+144`

`f(t)=80`

Hence, after 2 seconds the object reach its maximum height of 80 feet.