A particle is projected vertically upwards from the ground with speed \(30ms^{-1}\). Calculate the : (a) maximum height reached by the particle; (b) time ta...
Question 1 Report
A particle is projected vertically upwards from the ground with speed \(30ms^{-1}\). Calculate the :
(a) maximum height reached by the particle;
(b) time taken by the particle to return to the ground;
(c) time(s) taken for the particle to attain a height of 40m above the ground. [Take \(g = 10ms^{-2}\)]
Answer Details
Given \(u = 30\,\text{ms}^{-1},\ g = 10\,\text{ms}^{-2}\), upward taken positive.
(a) At maximum height \(v = 0\). Use \(v^2 = u^2 - 2gh\):
\[0 = 30^2 - 2(10)h \ \Rightarrow\ 20h = 900 \ \Rightarrow\ h = 45\,\text{m}\](b) Time to return to ground:
\[t = \frac{2u}{g} = \frac{2(30)}{10} = 6\,\text{s}\](c) Height \(40\,\text{m}\): use \(h = ut - \tfrac{1}{2}gt^2\):
\[40 = 30t - 5t^2 \ \Rightarrow\ 5t^2 - 30t + 40 = 0 \ \Rightarrow\ t^2 - 6t + 8 = 0\] \[(t - 2)(t - 4) = 0 \ \Rightarrow\ t = 2\,\text{s (going up)},\ t = 4\,\text{s (coming down)}\]Download The App On Google Playstore
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