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Answer 1 · 2017-04-06T18:59:09

Use the fact that repeatedly being #75%# of something is the same as repeatedly multiplying by #0.75#

a)

The first bounce is:

#(0.75)(100" meters") = 75" meters"#

The second bounce is the first bounce multiplied by #0.75#:

#(0.75)^2(100" meters") = 56.3" meters"#

The third bounce is the second bounce multiplied by #0.75#:

#(0.75)^3(100" meters") = 42.2" meters"#

b)
The nth bounce is:

#b(n) = (0.75)^n(100" meters")#

The sixth bounce is:

#b(6) = (0.75)^6(100" meters")#

#b(6) = 17.8" meters"#

Answer 2 · 2017-04-06T19:05:16

See below.

Calling #gamma = 0.75# and #h_0=100#

we have

#h_k=gamma h_(k-1)# or

#h_k=gamma^k h_0#

The three first sequence terms are

#h_0,h_1,h_2# or

#h_0,h_0 gamma, h_0 gamma^2, h_0 gamma^3# or

#100,75,56.2,42.2#

after the sixth bounce we have

#h_6=h_0 gamma^6 = 100 xx 0.75^6 = 17.8#[m]