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-16tē
1)what is the maximum height reached by
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-> 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-16tē
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-16tē
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(52780) (Show Source):
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.
