Question

A beach ball rolls off a cliff and onto the beach. The height, in feet, of the beach ball can be modeled by the function h(t)=64–16t^2, where t represents time, in seconds.

What is the average rate of change in the height, in feet per second, during the first 1.25 seconds that the beach ball is in the air?

Answer 1

You are given h(t) = 64–16t^2.

You need to find h(1.25).

Set up:

h(1.25) = 64 - 16(1.25)^2

Take it from here.

Answer 2

Answer:

Step-by-step explanation:

Given the height of a beach ball rolling off a cliff as a function of time

h = 64 — 16t²

We want to find the average change in height when t = 1.25s

The rate of change of height is the differentiation of the height I.e

∆h/∆t → dh/dt

Therefore

h = 64 — 16t².

Therefore

Differential of a constant is zero

dh/dt = —32t

At t = 1.25s

dh/dt = —32(1.25)

dh/dt = —40 ft/s

So, the height is reducing at a rate of 40ft in every seconds.