A stone is dropped vertically downwards from the top of a tower of height 45m with a speed of 20 ms\(^{-1}\). Find the :
(a) time it takes to reach the ground ;
(b) speed with which it hits the ground. [Take \(g = 10 ms^{-2}\)].
Explanation
\(s = ut + \frac{1}{2} gt^{2}\)
\(45 = 20t + \frac{1}{2} (10)(t^{2})\)
\(45 = 20t + 5t^{2}\)
\(5t^{2} + 20t - 45 = 0\)
\(\equiv t^{2} + 4t - 9 = 0\)
= \(\frac{-4 \pm \sqrt{(4^{2}) - 4(1)(-9)}}{2(1)}\)
= \(\frac{-4 \pm \sqrt{16 + 36}}{2}\)
= \(\frac{-4 \pm 2\sqrt{13}}{2}\)
= \(\frac{-2 \pm \sqrt{13}\)
\(t = -2 + \sqrt{13} \text{ or } t = -2 - \sqrt{13}\)
The time cannot be negative therefore, \(t = - 2 + \sqrt{13} = 1.61s\).
(b) \(v^{2} = u^{2} + 2gs\)
\(v^{2} = 20^{2} + 2(10)(45)\)
\(v^{2} = 400 + 900\)
\(v^{2} = 1300\)
\(v = \sqrt{1300}\)
\(v = \sqrt{100 \times 13} = 10\sqrt{13} ms^{-1}\)