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 ti...
Question 1 Report
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.
Answer Details
Height: \(h = 20t - \tfrac{3}{2}t^2 - \tfrac{2}{3}t^3\). Velocity is \(v = \dfrac{dh}{dt}\):
\[v = 20 - 3t - 2t^2\](a) Momentarily at rest when \(v = 0\):
\[2t^2 + 3t - 20 = 0 \ \Rightarrow\ t = \frac{-3 \pm \sqrt{9 + 160}}{4} = \frac{-3 \pm 13}{4}\]Taking the positive root, \(t = \dfrac{10}{4} = 2.5\,\text{s}\).
(b) Maximum height at \(t = 2.5\,\text{s}\):
\[h = 20(2.5) - \tfrac{3}{2}(2.5)^2 - \tfrac{2}{3}(2.5)^3\] \[= 50 - \tfrac{3}{2}(6.25) - \tfrac{2}{3}(15.625) = 50 - 9.375 - 10.4167\] \[= 30.21\,\text{m}\ (\text{to 2 decimal places})\]Download The App On Google Playstore
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