Asked by Glen
A ball is thrown vertically into the air so that its height s after t seconds is given by the equation, s=-1/27(t^2)+4√t. Find its maximum height.
Answers
Answered by
Steve
ds/dt = -2/27 t + 2/√t
= (54-2t^(3/2))/(27√t)
ds/dt=0 when
54 - 2t^(3/2) = 0
t^(3/2) = 27
t = 9
now just find s(9)
= (54-2t^(3/2))/(27√t)
ds/dt=0 when
54 - 2t^(3/2) = 0
t^(3/2) = 27
t = 9
now just find s(9)
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