SOLUTION: A projectile is thrown upward so that its distance above the ground after t seconds is h = -16t2 + 440t. After how many seconds does it reach its maximum height?

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Question 1100940: A projectile is thrown upward so that its distance above the ground after t seconds is h = -16t2 + 440t.
After how many seconds does it reach its maximum height?
Found 4 solutions by Shin123, ikleyn, amalm06, Alan3354:
Answer by Shin123(626) About Me  (Show Source):
You can put this solution on YOUR website!
 The maximum height is the vertex of the parabola h=-16t2+440t which is -b/2a. So the max height is -440/-32= 13.75 seconds.

Answer by ikleyn(52754) About Me  (Show Source):
You can put this solution on YOUR website!
.
Your "height" function is quadratic

h(t) = -16*t^2 + 440*t.


And the height is maximal when this quadratic function gets its maximum value.


The general theory says:

    A quadratic function  ax^2 + bx + c  gets its maximum/minimum value at  x = -b%2F%282a%29.


So, substitute your values a= -16  and  b= 440 into the formula, and you will obtain that your quadratic "height" function gets its maximum at 

t = -440%2F%282%2A%28-16%29%29 = 13.75 seconds.

Solved.

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My other lessons in this site on finding the maximum/minimum of a quadratic function are
    - 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

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic "Finding minimum/maximum of quadratic functions".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.



Answer by amalm06(224) About Me  (Show Source):
You can put this solution on YOUR website!
Position=h=-16t2 + 440t
Velocity=dh/dt=-32t+440
At its maximum height, the velocity of the projectile is zero:
-32t+440=0
440=32t
t=13.75 s (Answer)

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A projectile is thrown upward so that its distance above the ground after t seconds is h = -16t2 + 440t.
After how many seconds does it reach its maximum height?
================
Without using derivatives:
---
The max height is the vertex of the parabola, at t = -b/2a
t = -440/-32
t = 13.75 seconds