# Distance of Falling Object

• Feb 2nd 2010, 09:55 AM
sologuitar
Distance of Falling Object
The distance traveled by a falling object (subject only to gravity, with wind resistance ignored) is directly proportional to the square of the time it takes to fall that far. If an object falls 100 feet in 2.5 seconds, how far does it fall in 5 seconds?

MY WORK:

I developed the equation d = k/(t^2) from the given information, where d = distance, k is the constant of variation and t = time in seconds.

After doing the math, I got d = 25 feet, which makes no sense. The right answer is 400 feet. How do I get 400 feet? Is my equation wrong?
• Feb 2nd 2010, 10:25 AM
running-gag
The distance is proportional to the square of the time
d = ktē and not k/tē
• Feb 2nd 2010, 02:26 PM
sologuitar
ok...
Quote:

Originally Posted by running-gag
The distance is proportional to the square of the time
d = ktē and not k/tē

Originally, I set up the equation as d = k/(t^2) because the word proportional directed me to the fraction normally used when setting up a proportion.

100 = k(2.5)^2

100 = k(6.25)

100/6.25 = k

16 = k

Then d = 16(5)^2

d = 16 * 25

d = 400 feet

Is this right?

• Feb 3rd 2010, 12:22 PM
running-gag
Yes it is (Nod)

Note that you do not have to calculate the value of k
d = k(5)^2
100 = k(2.5)^2

Dividing
d/100 = 5^2/2.5^2
d/100 = 4
d = 400