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 = \(10 ms^{-2}\)].
(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
\(v = u + at = u - gt\)
\(0 = 80 - 10t \implies 80 = 10t\)
\(t = 8s\)
But time to return to projection plane = twice the time taken to reach maximum height = 2 x 8 = 16 secs.
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