height of a ball drop

Question

a certain ball is dropped from a height of x feet. it always bounces up to (2/3)x feet. suppose the ball is dropped from 10 feet and is stopped exactly when it touches the ground after the 30th bounce. What is the total distance travelled by the bal

Answer 1

Hi Cari!

 

It sometimes helps me to set up in words what is happening, and then convert those words into formulas, until I see a pattern.

 

In this case, the ball is starting at a height of 10 feet, which it drops, and then bounces back up by (2/3)(10 feet).  At this point, the distance it has covered (in feet) is:

 

d = 10 + (2/3)(10)

 

Then it has to drop back to the ground to bounce again, falling through a distance equal to what it rose after the first bounce. So now the distance it has covered is:

 

d = 10 + (2/3)(10) + (2/3)(10) = 10 + (2)(2/3)(10)

 

Then it bounces back up to a height equal to 2/3 of its height on the previous bounce.  Since the previous bounce was to a height of (2/3)(10), this bounce goes to a height of (2/3)[(2/3)(10)] = (2/3)²(10).  It then falls through an equal distance back to the ground, so that, after the second bounce, the total distance covered is:

 

d = 10 + (2)(2/3)(10) + (2)(2/3)²(10)

 

We can make a similar argument for all the following bounces.  So now, there is a pattern emerging for the distance added by each bounce.  For the n^th bounce, we can thus say that the distance added will be (2)(2/3)ⁿ(10).  The factor of 2 out front accounts for the ball having to go up and down an equal distance after each bounce to get to the next bounce.

 

So, it turns out that, after the first term, this becomes a geometric sequence, because each successive term is just 2/3 times the previous term.  We can write the whole distance in a standard geometric form as:

 

d = 10 + Σ_i=1ⁿ a_i = 10 + Σ_i=1ⁿ rⁱa

 

where

 

a = the common base in all terms

r  = the multiplicative factor connecting successive terms

 

For our case, a = (2)(10) = 20, since (2)(10) appears in every term, and r=(2/3).  Thus, the first term would be (2/3)¹(2)(10), which is just what we have above.

 

Once we define the common base and the multiplicative factor of a geometric sequence, we can use the fact that the sum of the first n terms of the sequence is given by:

 

sum of first n factors = a[(1-rⁿ)/(1-r)]

 

Since n represents the number of bounces, we want to sum the sequence to n = 30.  That sum can be written as:

 

sum = 20[(1-(2/3)³⁰)/(1-(2/3))] 

 

Now, (2/3)³⁰ is a pretty small number, so the numerator inside the square brackets is essentially 1.  Thus, it becomes

 

sum = 20(1/(1-(2/3))) = 20(1/(1/3)) = 20(3) = 60 feet.

 

Now, we are almost there.  But there was also that first 10-foot drop before the first bounce to add in.  So the total distance d, in feet, is:

 

d = 10 + sum = 10 + 60 = 70 feet

 

If you want to go over any of these steps or algebra operations in more detail, just let me know.  I hope this helps!

 

 

Answer 2

X is 10 ft. So it drops 10 feet and then 2/3 * 30 * 10  or 10+200= 210 ft.