SOLUTION: An object is thrown straight up into the air then follows a trajectory , the height s(t)of the object is given by the function s(t)=4t-16t² 1)what is the maximum height reached by

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Question 1099516: An object is thrown straight up into the air then follows a trajectory , the height s(t)of the object is given by the function s(t)=4t-16t²
1)what is the maximum height reached by the object?
2)how long did the object reached its maximum height?
This is trajectory topic

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
It is about finding the vertex (the maximum) of the quadratic function

s(t) = -16*t^2 + 4t.


In the given case you can present the function as the product

s(t) = -4t*(4t-1)


of two factors -4t and (4t-1). Then it is clear that the quadratic function has the roots at t = 0  and t = 1/4.


Then the midpoint t = 1/8 is the point where the quadratic function reaches its maximum.
So the time to get maximum height is 1/8 of a second.

Then the maximum height is   =  = .


Answer.  The maximum height is  of the foot and it reaches at t = 1/8 of a second.




Plot h(t) =


There is another way to analyze the problem.

It is described in lessons
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower
in this site.

In any case, you need to know how to find the maximum/minimum of a quadratic function presented in the general form.

You will find it in 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.


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