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OpenStudy (anonymous):
an object is launched upward at 45ft/sec from a platform that is 40 feet high.what is the obhect maximum height if the equation of height (h) in terms of time(t) of the object given by h(t) =-16y^2 + 45t + 40?
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OpenStudy (mertsj):
Did you mean for the y^2 to be t^2?
OpenStudy (anonymous):
im sorry its suppose to be t^2
OpenStudy (mertsj):
Just find the y coordinate of the vertex of your equation. That will be the maximum height
OpenStudy (anonymous):
not quite understanding
OpenStudy (mertsj):
Do you know that the equation you posted is the equation of a parabola?
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OpenStudy (anonymous):
no I didn't
OpenStudy (mertsj):
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