SOLUTION: I am having difficulties on how to work this type of word problem. any help appreciated. The height s, in feet, of a rock thrown upward at an initial speed of 64ft/sec from a

Question 165020This question is from textbook
:
I am having difficulties on how to work this type of word problem. any help appreciated.
The height s, in feet, of a rock thrown upward at an initial speed of 64ft/sec from a cliff 50 ft. above an ocean beach is given by the function
s(t) = -16^2 + 64t +50, where t is the time in seconds. Find the maximum height above the beach that the rock will attain. This question is from textbook

Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Ref:
http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php
.
The quadratic is a parabola:
f(x) = ax^2 + bx + c
.
If the coefficient 'a' is negative (which is your case), the parabola opens downward (picture -- "negative" -- sad face).
.
Therefore, the "vertex" of the parabola will give you the "maximum".
.
Vertex form of a parabola:
y= a(x-h)^2+k
where
(h, k) is the vertex
.
The idea is to convert your given equation into the "vertex" form.
s(t) = -16t^2 + 64t +50
Group terms:
s(t) = (-16^2 + 64t) +50
s(t) = -16(t^2 - 4t) +50
"Complete the square":
see:http://www.purplemath.com/modules/sqrquad.htm
.
s(t) = -16(t^2 - 4t + 4) +50 + 4(16)
s(t) = -16(t^2 - 4t + 4) +50 + 64
s(t) = -16(t^2 - 4t + 4) + 114
s(t) = -16(t - 2)^2 + 114
.
Vertex:
(h,k)= (2,114)
This says, that at 2 seconds, the rock is at 114 feet.
.
Answer: 114 feet

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