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.
(a) the time at which it is momentarily at rest (b) correct to two decimal places, the maximum height reached by the object.
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Correct Answer: Option n
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 \(t = \frac{5}{2}s\).
\(h_{m} = 20(\frac{5}{2}) - \frac{3}{2}(\frac{5}{2})^{2} - \frac{2}{3}(\frac{5}{2})^{3}\)
= \(50 - \frac{75}{8} - \frac{250}{16} = \frac{400}{16} = 25 m\)
\(h = 20t - \frac{3}{2}t^{2} - \frac{2}{3}t^{3}\)
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 \(t = \frac{5}{2}s\).
\(h_{m} = 20(\frac{5}{2}) - \frac{3}{2}(\frac{5}{2})^{2} - \frac{2}{3}(\frac{5}{2})^{3}\)
= \(50 - \frac{75}{8} - \frac{250}{16} = \frac{400}{16} = 25 m\)