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Mathematics 18 Online
OpenStudy (anonymous):

An object is launched at 19.6 m/s from a height of 58.8 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.9t^2 +19.6t + 58.8. What is the object's maximum height? A. 58.8 B. 194.04 C. 78.4 D. 117.6

OpenStudy (lieutenantgeneral):

-4.9t^2 +19.6t + 58.8 = -4.9 (t - 6.) (t + 2.) -4.9 (t - 6.) (t + 2.) t (19.6 - 4.9 t) + 58.8 t = -2 t = 6

OpenStudy (anonymous):

then what ? plug it in ?

OpenStudy (anonymous):

@AravindG

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