A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).
Calculate the maximum height reached.
S = 12t + \(\frac{5}{2t^2}\) - t\(^3\);
ds/dt = 12 + 5t - 3t\(^2\)
At max height ds/dt = 0
i.e 12 + 5t - 3t\(^2\)
(3t + 4)(t -3) = 0;
t = -4/3 or 3
Hmax = 12[3] + \(\frac{5}{2[3]^2}\) - 3\(^3\)
= 36 + 45/2 - 27
= 31.5m
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