An object is thrown vertically upwards from the top of a cliff with a velocity of \(25ms^{-1}\). Find the time, in seconds, when it is 20 metres above the cliff. \([g = 10ms^{-2}]\).
\(s = ut + \frac{at^{2}}{2}\)
This movement is against gravity, so it is negative.
\(s = ut - \frac{gt^{2}}{2}\)
\(s = 20m, u = 25ms^{-1}\)
\(20 = 25t - \frac{10t^{2}}{2} \implies 20 = 25t - 5t^{2}\)
\(5t^{2} - 25t + 20 = 0 \)
\(5t^{2} - 5t - 20t + 20 = 0 \implies 5t(t - 1) - 20(t - 1) = 0\)
\(5t - 20 = \text{0 or t - 1 = 0}\)
\(t = \text{1 or 4}\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}