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)