The distance that a free falling object falls is directly proportional to the square of the time it falls

Question

The distance that a free falling object falls is directly proportional to the square of the time it falls (before it hits the ground). If an object fell 44 ft in 5 seconds, how far will it have fallen by the end of 6 seconds? (Leave the variation constant in fraction form or round to at least 2 decimal places. Round your final answer to the nearest foot.)

Answer 1

Let's first come up with some variables, and then perform an English-to-Math translation to derive an equation.

 

d = distance an object falls

t = the time it falls

 

Imagine the object in hand, and t is measured by a stopwatch.  The stopwatch starts at the moment the object is released; at this moment, t = 0.  d is also equal to 0 when t = 0, since at the precise moment it is released, it hasn't fallen yet.

 

Now that we have the situation and variables defined, we can begin the English-to-Math translation:

 

"The distance a free falling object falls"  "is" "directly proportional to" "the square of the time it falls"

                         d                                =              k                                      t²

 

Removing all of the blank space, we have

 

d = kt².

 

We are also given a distance (d = 44) for a given time (t = 5).  We can solve for k by substituting these values into our equation:

 

44 = k (5²).  Since 5² = 25,

44 = 25k.  Divide both sides by 25:

44/25 = k.

 

So, our equation is now complete:

 

d = (44/25) * t².

 

"How far will it have fallen in 6 seconds" means, what is d if t = 6?  This is equivalent to putting 6 in place of t:

 

d = (44/25) * 6², or

d = (44/25) * 36.

 

Do the calculation by hand or calculator, and you have your answer.