SOLUTION: A ball is thrown upward such that it's height (h) in t seconds is given by h=-16t^2 +8t +48 What is the maximum height reached by the ball?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A ball is thrown upward such that it's height (h) in t seconds is given by h=-16t^2 +8t +48 What is the maximum height reached by the ball?      Log On


   

Question 1054211: A ball is thrown upward such that it's height (h) in t seconds is given by
h=-16t^2 +8t +48
What is the maximum height reached by the ball?
Found 2 solutions by Fombitz, ikleyn:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Convert to vertex form by completing the square,
h=-16t%5E2%2B8t%2B48

h=-16%28t-1%2F4%29%5E2%2B49
So the max height of 49 occurs when t=1%2F4s.

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
A ball is thrown upward such that it's height (h) in t seconds is given by
h=-16t^2 +8t +48
What is the maximum height reached by the ball?
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Solution 

Apply completing the square to find the maximum of the given quadratic function.


-16t%5E2+%2B+8t+%2B48 = -%284t-1%29%5E2+%2B1+%2B+48 = -%284t-1%29%5E2+%2B+49.


The maximum is 49 feet.

See the lessons
    - HOW TO complete the square to find the minimum/maximum of a quadratic function
    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola
in this site.

See also the lessons
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower