Tutoring high school physics, this topic is important.  The tutor introduces it.

A problem that might be seen by a grade 11 physics student is the following:

Imagine Isabel is reading a book in the sun on her second-floor backyard deck. She leaves the book on the rail to take a call. Forgetting about it, she elbows the book off the railing.  Assuming it’s 5.5m above the ground, how long before the book hits the lawn?

Solution: A formula for the distance covered by an accelerating object is

d=v₀t + 0.5at^2

where

v₀ is the starting velocity
a is the acceleration
t is the time

In Isabel’s case, we know the distance is -5.5m (negative because it’s downward). We know the original velocity of the book is 0, since it’s at rest before it starts falling. Furthermore, we know its acceleration is -9.8m/s², the acceleration due to gravity (on Earth).

We can plug these values into the formula:

-5.5=0(t) + 0.5(-9.8)t^2

Simplifying, we get

-5.5=-4.9t^2

We need to solve for t:

-5.5/-4.9=t^2

Dividing, we get

1.122=t^2

Finally, we square root both sides:

1.059=t

So, in significant digits, t=1.1s. It takes 1.1s for the book to fall.

Don’t worry; I’m sure it was a paperback:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.