A 20-foot ladder is leaning against a vertical wall. The bottom of the ladder is pulled away horizontally from the wall at 3 feet per second. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 10 feet away?

This question can be divided into two different equations. In order to calculate how fast the ladder is sliding downwards, you need to know at what original vertical height the ladder is leaning on the wall. Then you can calculate by how far the ladder is sliding down vertically if it is pulled away horizontally from the wall by 3 feet per second.Firstly, the original vertical intersection of the ladder with the wall can be calculated with Pythagoras, saying that the sum of the squares of the length of the wall and the bottom equals the length of the ladder (a²+b²=c²). By simply inserting the value of 10 feet into a, and inserting the length of the ladder into c, the value can easily be calculated by solving for b and taking the square roots. Hereby, b equals to 17.32 .The same procedure should be done for the second part of the question, assuming now that the bottom length is now 13 feet, instead of 10 feet. Subbing the value of 13 into a now, and the length of the ladder (which still is 20) into c, b should equal to appr. 15.198.The difference between 17.32 and 15.198 is the length that the ladder moves down per second vertically when you pull the ladder 3 feet per second away horizontally from the wall. Hence, the ladder is sliding down by 17.32-15.198= 2.122 feet per second.