A particle is projected vertically upwards from the ground with speed \(30ms^{-1}\). Calculate the :...
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 \(g = 10ms^{-2}\)]
Explanation
v = 0 m/s ; u = 30 m/s ; s = ? ; g = 10 m/s\(^{2}\).
(a) For maximum height, we use the formula
\(v^{2} = u^{2} - 2gs\)
\(s = \frac{u^{2} - v^{2}}{2g}\)
= \(\frac{30^{2} - 0^{2}}{2(10)}\)
= \(\frac{900}{20} = 45m\)
(b) Time to reach highest point:
\(v = u - gt \implies t = \frac{u - v}{g}\)
\(t = \frac{30 - 0}{10} = 3s\)
If the particle took 3s to get to the maximum height,
Time to reach the ground = 2(3) = 6s.
(c) From a height 40m above the ground, we use
\(s = ut - \frac{1}{2} gt^{2}\)
\(40 = 30t - 5t^{2}\)
\(5t^{2} - 30t + 40 = 0\)
\(5t^{2} - 20t - 10t + 40 = 0 \implies 5t(t - 4) - 10(t - 4) = 0\)
\((5t - 10)(t - 4) = 0 \implies \text{2s and 4s}\)
\(\therefore\) At times 2s and 4s, the height attained = 40m.