Question

An object is launched directly in the air at a speed of 88 feet per second from a platform located 4 feet in the air. The motion of the object can be modeled using the function f(t)=−16t^2+88t+4, where t is the time in seconds and f(t) is the height, in feet, of the object. What is the maximum height, in feet, that the object will reach? Do not include units in your answer.

Answer 1

Answer:

The maximum height that the object will reach is 125 feet.

Explanation:

Hi there!

Let´s write the function of height:

f(t) = -16 · t² + 88 · t + 4

Notice that the function is a negative parabola. The maximum of the function is located at the vertex of the parabola. At the vertex, the line tangent to that point is horizontal, i.e., its slope is zero. The slope of the tangent line is the derivative of the function and it also expresses the velocity of the object:

df/dt = -32 · t + 88

At the maximum point, the slope of the tangent line is zero (it is the same as to say that at the maximum height, the velocity of the object is zero), then:

df/dt = 0

0 = -32 · t + 88

Solving the equation for t:

-88/-32 = t

t = 2.75 s

Then, the maximum height will be f(2.75):

f(t) = -16 · t² + 88 · t + 4

f(2.75) = -16(2.75)² + 88(2.75) + 4

f(2.75) = 125 ft

The maximum height that the object will reach is 125 feet.