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.
Explanation
\(h = 20t - \frac{3}{2}t^{2} - \frac{2}{3}t^{3}\) 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 \(t = \frac{5}{2}s\). \(h_{m} = 20(\frac{5}{2}) - \frac{3}{2}(\frac{5}{2})^{2} - \frac{2}{3}(\frac{5}{2})^{3}\) = \(50 - \frac{75}{8} - \frac{250}{16} = \frac{400}{16} = 25 m\)
