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.