Pythagorean Theorem

Question

A ladder 25 feet long is leaning against a house. At time t = 0, the base of the ladder is 7 feet from the wall, and the base of the ladder starts to move away from the wall at a constant rate of 2 feet per second.

a. How far is the top of the ladder from the ground at t = 0?

b. How far is the base of the ladder from the wall and how far is the top of the ladder from the ground at t = 4?

c. At t = 8.5 seconds, how far is the base of the ladder from the wall and how far is the top of the ladder from the ground?

Answer 1

The Pythagorean Theorem states that a²+b²=c².

Also, for a constant rate, displacement = speed * time

a) This first question is not a calculus question, so we can use the Pythagorean Theorem:

7² + b² = 25²

b² = 576

b = 24 feet (answer, height of ladder on wall)

b) Since we are studying the time when t = 4, we know that the base of the ladder has moved (2 feet/sec)(4 sec) = 8 feet. Since the base of the ladder was already 7 feet from the house at the beginning, we know that the base of the ladder is now 15 feet from the house.

15 feet (first answer, distance from base of ladder to house)

The ladder is (of course) still 25 feet long, so we can write:

15² + b² = 25²

225 + b² = 625

b² = 400

b = 20 (second answer, height of ladder on wall)

c) By the time t = 8.5, the ladder has moved even further from the starting position of 7 feet. Using displacement = speed * time, we get (2 ft/sec)(8.5 sec) = 17 feet

So now the base of the ladder is 7 + 17:

24 feet (answer, distance from base of ladder to house)

Using the Pythagorean Theorem one last time, we have:

24² + b² = 25²

576 + b² = 625

b² = 49

b = 7 feet (answer, height of ladder on wall)