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 2sqrt{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} = 10sqrt{13} ms^{-1})