An object is projected vertically upwards with a velocity of 80 m/s. Find the :
(a) Maximum height reached
(b) Time taken to return to the point of projection. [Take g = (10 ms^{-2})].
Explanation
(a) Maximum height reached
Using the equation, (v^{2} = u^{2} – 2gs) (Upward movement against gravity)
velocity at maximum height = 0 m/s
(0 = 80^{2} – 2(10)s)
(0 = 6400 – 20s implies 6400 = 20s )
(s = 320 m)
(b) Time to reach the maximum height
(v = u + at = u – gt)
(0 = 80 – 10t implies 80 = 10t)
(t = 8s)