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Mathematics 5 Online
OpenStudy (anonymous):

H(t)=-16t^2+vt+s what is maximum height? I forgot to say that the initial velocity is 34 per second

OpenStudy (anonymous):

For the maximum height recall that your velocity = 0 so take the derivate of that function for velocity. \[\frac{ dH }{ dt } = -32t + v\] make the derivate = 0 and solve for time sub back the time in the max height equation. H(t).

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