extreme values?

Question

A stone is tossed from a bridge into the river below. h= -2t^2+7t +9 A) find the maximum height of the stone B)  how many seconds does it take to reach the maximum height? C) how many seconds does it take for the stone to reach the water?

Answer 1

h=-(2t^2-7t-9)=-(2t-9)(t+1)=(9-2t)(t+1); h'=-4t+7, which is maximum at t=7/4 (h'=0). So maximum height=(11/2)(11/4)=121/8=15.125. It takes 7/4=1.75 seconds to reach maximum height. If h is the height above the water, then the stone reaches the water when h=0, 9-2t=0, t=4.5 seconds.

Alternative solution for first part (without using calculus):

h=-2(t^2-(7/2)t-9/2)=-2(t^2-(7/2)t+(7/4)^2-72/16-(7/4)^2)=-2((t-7/4)^2-121/16)=-2((t-7/4)^2-(11/4)^2) (completing the square).

So h=2((11/4)^2-(t-7/4)^2), and h is clearly maximum when (t-7/4)^2=0, its minimum value, when t=7/4=1.75 seconds and h=2*121/16=121/8=15.125.