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Mathematics 13 Online
OpenStudy (anonymous):

An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.9t2 + 20t + 65. What is the object's maximum height?

OpenStudy (anonymous):

are you learning about derivatives or parabolas (vertex formula) ?

OpenStudy (anonymous):

parabolas .... I got 85m but can you help me step by step so I can get it stuck in my head? haha

OpenStudy (anonymous):

(x,y) = (-b/2a , (4ac-b^2)/4a) those are the coord.s of the vertex...

OpenStudy (anonymous):

so (4*-4.9*65 - 20^2)/(4*-4.9)

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