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...
Question 1 Report
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 = \(10 ms^{-2}\)].
Answer Details
Take upward as positive, initial velocity \(u = 80\ \text{m/s}\), \(g = 10\ \text{m/s}^2\).
(a) Maximum height. At the highest point the velocity is zero. Using \(v^2 = u^2 - 2gH\) with \(v = 0\):
\[H = \frac{u^2}{2g} = \frac{80^2}{2\times 10} = \frac{6400}{20} = 320\ \text{m}.\]The maximum height reached is 320 m.
(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 \(T = \dfrac{2u}{g} = \dfrac{160}{10} = 16\ \text{s}\).)
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