A ball is dropped from an initial height of 80 feet. After each time it hits the ground, it rebounds...

1. TRANSLATE the problem information

• Given information:
  • Initial height: 80 feet
  • After each bounce, height becomes 50% (or 0.5 times) the previous height
  • h = height after nth bounce
  • Need formula for h in terms of n

2. INFER the mathematical pattern

• This describes a geometric sequence where each term is 0.5 times the previous term
• Starting value is what happens after the first bounce: \(\mathrm{80 \times 0.5 = 40}\) feet
• Common ratio is 0.5
• We need to determine if this follows \(\mathrm{h = 80(0.5)^n}\) or a different exponential pattern

3. VISUALIZE the first few bounces

Let me trace what happens:

• After 1st bounce \(\mathrm{(n=1)}\): \(\mathrm{h_1 = 80 \times 0.5 = 40}\) feet
• After 2nd bounce \(\mathrm{(n=2)}\): \(\mathrm{h_2 = 40 \times 0.5 = 80 \times (0.5)^2 = 20}\) feet
• After 3rd bounce \(\mathrm{(n=3)}\): \(\mathrm{h_3 = 20 \times 0.5 = 80 \times (0.5)^3 = 10}\) feet

4. INFER the general formula

• Pattern shows: \(\mathrm{h = 80(0.5)^n}\)
• The 80 represents the original height
• The \(\mathrm{(0.5)^n}\) represents the decay factor applied n times

5. SIMPLIFY by testing answer choices

Test with \(\mathrm{n = 1}\) (should give \(\mathrm{h = 40}\)):

• (A) \(\mathrm{h = 40(0.5)^1 = 20}\) ✗
• (B) \(\mathrm{h = 80(0.5)^{1-1} = 80(1) = 80}\) ✗
• (C) \(\mathrm{h = 80(0.5)^1 = 40}\) ✓
• (D) \(\mathrm{h = 80(2)^1 = 160}\) ✗

Verify with \(\mathrm{n = 2}\) (should give \(\mathrm{h = 20}\)):

• (C) \(\mathrm{h = 80(0.5)^2 = 80(0.25) = 20}\) ✓

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret what "after its nth bounce" means, thinking \(\mathrm{n=1}\) should give the original 80 feet instead of the height after the first bounce (40 feet).

This leads them to look for a formula that gives 80 when \(\mathrm{n=1}\), making them select Choice B (\(\mathrm{h = 80(0.5)^{n-1}}\)) because \(\mathrm{80(0.5)^0 = 80}\).

Second Most Common Error:

Poor INFER reasoning: Students recognize the 50% decay but incorrectly think the formula should start with 40 (the first bounce height) rather than 80 (the original height).

This makes them select Choice A (\(\mathrm{h = 40(0.5)^n}\)) because it starts with 40, but they don't verify that this gives wrong values for subsequent bounces.

The Bottom Line:

This problem requires careful attention to the indexing - understanding that n represents the number of bounces that have occurred, not the drop number, and that the formula must account for the original height being reduced by the decay factor n times.