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 projecti...
Assessment:WAEC SSCE - Further Mathematics - 2006 (Essay)Subject:Further Mathematics
(b) Time to return to the point of projection. By symmetry the total time is twice the time to reach the top. Time up: \(v = u - gt \Rightarrow 0 = 80 - 10t \Rightarrow t = 8\ \text{s}\).
\[T = 2t = 2\times 8 = 16\ \text{s}.\]
The object returns after 16 s. (Equivalently \(T = \dfrac{2u}{g} = \dfrac{160}{10} = 16\ \text{s}\).)
(b) Time to return to the point of projection. By symmetry the total time is twice the time to reach the top. Time up: \(v = u - gt \Rightarrow 0 = 80 - 10t \Rightarrow t = 8\ \text{s}\).
\[T = 2t = 2\times 8 = 16\ \text{s}.\]
The object returns after 16 s. (Equivalently \(T = \dfrac{2u}{g} = \dfrac{160}{10} = 16\ \text{s}\).)