1963 AHSME Problems/Problem 29

Problem

A particle projected vertically upward reaches, at the end of $t$ seconds, an elevation of $s$ feet where $s = 160 t - 16t^2$ . The highest elevation is:

$\textbf{(A)}\ 800 \qquad \textbf{(B)}\ 640\qquad \textbf{(C)}\ 400 \qquad \textbf{(D)}\ 320 \qquad \textbf{(E)}\ 160$

Solution

The highest elevation a particle can reach is the vertex of the quadratic. The x-value that can get the maximum is $\frac{-160}{-2 \cdot 16} = 5$ , so the highest elevation is $160(5) - 16(5^2) = 400$ feet, which is answer choice $\boxed{\textbf{(C)}}$ .

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
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1963 AHSME Problems/Problem 29

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