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 grou...
Question 1 Report
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 \(g = 10 ms^{-2}\)].
Answer Details
Take downward as positive with initial speed \(u = 20\text{ ms}^{-1}\), \(g = 10\text{ ms}^{-2}\), height \(s = 45\text{ m}\).
(a) Time to reach the ground. Using \( s = ut + \tfrac12 g t^2 \):
\[ 45 = 20t + 5t^2 \Rightarrow 5t^2 + 20t - 45 = 0 \Rightarrow t^2 + 4t - 9 = 0. \]
\[ t = \frac{-4 + \sqrt{16 + 36}}{2} = \frac{-4 + \sqrt{52}}{2} = \frac{-4 + 7.211}{2} \approx 1.61\text{ s}. \]
(b) Speed on hitting the ground. Using \( v^2 = u^2 + 2gs \):
\[ v^2 = 20^2 + 2(10)(45) = 400 + 900 = 1300 \Rightarrow v = \sqrt{1300} \approx 36.1\text{ ms}^{-1}. \]
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