Question

A ball is thrown off the top of a very tall building at time

t = 0. The height s(t) of the ball (above ground level) at time t is
given by the formula
-16² +50t + 1200.
What is the average velocity of the ball on the interval [1, 3/2]? That
is, what is the average velocity of the ball over the half-second pe-
riod starting exactly one second after the ball is thrown?

Answer 1

the average velocity of the ball on the interval [1, 3/2] is -808 ft/s.

Explanation:

The height of the ball at time t is given by the formula:

s(t) = -16t^2 + 50t + 1200

We need to find the average velocity of the ball on the interval [1, 3/2]. The average velocity is defined as the change in position divided by the change in time, or:

average velocity = (s(3/2) - s(1)) / (3/2 - 1)

Substituting the formula for s(t), we get:

average velocity = ((-16(3/2)^2 + 50(3/2) + 1200) - (-16(1)^2 + 50(1) + 1200)) / (3/2 - 1)

Simplifying and solving for the average velocity, we get:

average velocity = (430 - 1234) / (1/2) = -808 ft/s

Therefore, the average velocity of the ball on the interval [1, 3/2] is -808 ft/s.