Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
1. Section 5.1
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Exercise 34 Page 208

Practice makes perfect
a DeShawn and and her team gathered the information shown in the table below. We are asked to find the rebound rate for the ball they used.
c | c Drop Height & Rebound Height 150 cm & 124 cm 70 cm & 59 cm 120 cm & 100 cm 100 cm & 83 cm 110 cm & 92 cm 40 cm & 33 cm

We can calculate the rebound rate as the quotient between the drop height and the rebound height. However, in this case we have different measurements which would give us different values. To consider all of them, we can find the rebound rates for each pair of values and then average them. Average value = x_1+x_2+x_3 + ... + x_n/n In this formula, x_1+x_2+x_3 + ... + x_n represent the values to be averaged, and n is the number of values we have in total. If we do the calculations we find the results shown in the following table.

Drop Height H_d Rebound Height H_r Rebound Rate H_r/H_d
150 cm 124 cm 0.827
70 cm 59 cm 0.843
120 cm 100 cm 0.833
100 cm 83 cm 0.83
110 cm 92 cm 0.836
40 cm 33 cm 0.825
Average Rebound Rate: 0.832
b As we found in Part A, our best prediction for the rebound rate is 0.832. This means that the quotient between the rebound height H_r and the drop height H_d will be 0.832 for the ball used.

H_r/H_d= 0.832

We can use this to find the rebound height when the ball is dropped from 275 cm. We should just substitute the given value for the drop height and solve for the rebound height the ball will reach. We will be rounding our answer to the nearest cm according to the precision of the measurements.

H_r/H_d= 0.832
H_r/275= 0.832
H_r/275 * 275= 0.832 * 275
H_r = 0.832 * 275
H_r= 228.8
H_r= 229
Our prediction for the reached height is 229 cm.
c Now we will find the height from where ball was dropped, knowing that its rebound height was 60 cm. This time we need to substitute the value for the rebound height H_r into the formula and then solve for the drop height H_d.
H_r/H_d= 0.832
60/H_d= 0.832
â–Ľ
Solve for H_d
60/H_d * H_d= 0.832 * H_d
60 = 0.832H_d
60/0.832 = 0.832H_d/0.832
72.115384... = H_d
H_d = 72.115384 ...
H_d = 72
Once again, we rounded to the nearest integer as the measurements shown in the table were precise up to one centimeter. Our prediction for the drop height is 72 cm.
d For this part we need to predict the rebound height after the first bounce when dropping the ball from a window 200 meters up the Empire State Building. This is similar to what we did in Part B, so we will proceed in the same way.
H_r/200= 0.832
H_r/200= 0.832
â–Ľ
Solve for H_r
H_r/200 * 200= 0.832 * 200
H_r = 0.832 * 200
H_r= 166.4
H_r= 166
As we can see, the height reached after the first bounce would be 166 cm.

We have found that the rebound height the ball will reach after the first rebound is approximately 138 m.

e To find the height reached at the second and third bounce we have to do the same calculation we have done before recursively. This means we will be using the height of the previous bounce as the initial height.


Furthermore, notice that each time we do this calculation, all we are doing is multiplying the corresponding initial height by the rebound rate. If we do this recursively we can find the height after any number of bounces we want.

Bounce Rebound Height Rounded Answer (m)
1 200 * 0.832_1 166
2 200 * 0.832 * 0.832_2 138
3 200 * 0.832 * 0.832 * 0.832_3 115

We can see that for the second bounce the height reached would be 138 m, and 115 m for the third one.