1963 AHSME Problems/Problem 29

Revision as of 16:50, 7 June 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 29)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $\boxed{\textbf{(C)}}$.

See Also

1963 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 30
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS