A problem that might be seen by a grade 11 physics student is the following:<\/p>\n
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. \u00a0Assuming it’s 5.5m above the ground, how long before the book hits the lawn?<\/p>\n
Solution: A formula for the distance covered by an accelerating object is<\/p>\n
d=v0<\/sub>t + 0.5at^2<\/p>\n where<\/p>\n v0<\/sub> is the starting velocity 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<\/sup>, the acceleration due to gravity (on Earth).<\/p>\n We can plug these values into the formula:<\/p>\n -5.5=0(t) + 0.5(-9.8)t^2<\/p>\n Simplifying, we get<\/p>\n -5.5=-4.9t^2<\/p>\n We need to solve for t:<\/p>\n -5.5\/-4.9=t^2<\/p>\n Dividing, we get<\/p>\n 1.122=t^2<\/p>\n Finally, we square root both sides:<\/p>\n 1.059=t<\/p>\n So, in significant digits, t=1.1s. It takes 1.1s for the book to fall.<\/p>\n Don’t worry; I’m sure it was a paperback:)<\/p>\n
\na is the acceleration
\nt is the time<\/p>\n