Question

A bouncy ball is dropped such that the height of its first bounce is 6.25 feet and each

Answer 1

Answer:

0.67 feet

Step-by-step explanation:

This question can be interpreted as find the 11th term of a geometry progression where a = 6.25 and r = 80%

So, we have:

$T_n = ar^{n-1}$

In this case:

$r - 80\% = 0.8$

$n = 11$

So, we have:

$T_{11} = 6.25 * 0.8^{11-1}$

$T_{11} = 6.25 * 0.8^{10}$

$T_{11} = 6.25 * 0.1073741824$

$T_{11} = 0.67ft$

Hence, the height of rebound at the 11th time is approximately 0.67

Answer 2

Answer: 0.7

Step-by-step explanation: