Problem: A particle projected vertically upward reaches; at the end of t seconds, an elevation of s feet where s=160t−16t2. The highest elevation is:
Answer Choices:
A. 800
B. 640
C. 400
D. 320
E. 160
Solution:
The abscissa of the maximum (or minimum) point on the parabola ax2+bx+c is −b/2a. For the given parabola 160t−16t2,−b/2a=−160/−32=5, and the value of s when t=5 is 400.
or
Since s=160t−16t2,s=400−400+160t−16t2=400−16(5−t)2. The right side of this equation is a maximum when 5−t=0. Therefore, s(max)=400.