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

successive bounce is 80% of the previous bounce's height. What would be the height

of the 11th bounce of the ball?

Sagot :

MrRoyalidn

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:

[tex]T_n = ar^{n-1}[/tex]

In this case:

[tex]r - 80\% = 0.8[/tex]

[tex]n = 11[/tex]

So, we have:

[tex]T_{11} = 6.25 * 0.8^{11-1}[/tex]

[tex]T_{11} = 6.25 * 0.8^{10}[/tex]

[tex]T_{11} = 6.25 * 0.1073741824[/tex]

[tex]T_{11} = 0.67ft[/tex]

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

Answer: 0.7

Step-by-step explanation:

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