An object is projected vertically upwards. Its height, h m, at time t seconds is given by \(h = 20t - \frac{3}{2}t^{2} - \frac{2}{3}t^{3}\). Find
(a) the time at which it is momentarily at rest (b) correct to two decimal places, the maximum height reached by the object.
\(h = 20t - \frac{3}{2}t^{2} - \frac{2}{3}t^{3}\)
Velocity, \(\frac{\mathrm d h}{\mathrm d t} = 20 - 3t - 2t^{2}\)
At point of rest, v = 0
\(2t^{2} + 3t - 20 = 0\)
\(\implies 2t^{2} - 5t + 8t - 20 = 0\)
\(t(2t - 5) + 4(2t - 5) = 0 \implies t = \frac{5}{2}; -4\)
The value of time cannot be negative hence \(t = \frac{5}{2}secs\)
(b) Maximum height, \(h_{m}\) is at time \(t = \frac{5}{2}s\).
\(h_{m} = 20(\frac{5}{2}) - \frac{3}{2}(\frac{5}{2})^{2} - \frac{2}{3}(\frac{5}{2})^{3}\)
= \(50 - \frac{75}{8} - \frac{250}{16} = \frac{400}{16} = 25 m\)
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