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A basketball is thrown upward from a height of 1.8 m at a velocity of 12 m/s. How long until it reaches the ground?
(Recall: h(t)=-4.9t^2-v_0 t+h_0, where v_0 is initial velocity and h_0 is initial height.)

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Answer:

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Step-by-step explanation:

1 Lecture 15: Rates of change

1.1 Outline

• Rates of change, velocity, acceleration.

• Marginal rates

• Equation of motion for objects falling under constant gravity

• Interpreting position and velocity graphs

1.2 Rates of change

If f(t) is a function depending on time, then we can write down the average rate of

change on an interval [t1, t2],

f(t2) − f(t1)

t2 − t1

.

We have defined the instantaneous rate of change at a time t as the limit

f

0

(t) = limr→t

f(r) − f(t)

r − t

.

This is also called the derivative. Thus, instantaneous rate of change is another name

for the derivative. In applications the name rate of change is more descriptive.

If s(t) gives the position of a particle as it moves along a line, then

v(t) = limk→0

s(t + k) − s(t)

k

gives the instantaneous velocity. If position is measured in meters and time in seconds,

the velocity v is measured in meters/second. We will abbreviate meters by m and

seconds by s so that the units for velocity are m/s.

As a second example, let v(t) be a function which gives the v

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