1963 AMC 12 Problem 29

Attempt Problem 29 of the 1963 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1963 AMC 12 solutions, or check the answer key.

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29.

A particle projected vertically upward reaches, at the end of tt seconds, an elevation of ss feet where s=160t−16t2.s=160t-16t^2. The highest elevation is:
800800
640640
400400
320320
160160
Answer: C
Concepts:quadraticoptimization
Small hint:
The height is a downward-opening quadratic
Big hint:
Find its vertex time using −b2a-\frac{b}{2a}
Solution:
The vertex occurs at t=−1602⋅−16=5.t=-\frac{160}{2\cdot-16}=5. Then s(5)=160(5)−16(25)=400.s(5)=160(5)-16(25)=400.
Thus, the correct answer is C.
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