A Stone Falls From A Cliff And Travels 24.5 m In The Last Second Before It Reaches The Ground At The Foot Of Teh Clif. Find The Height Of The Cliff.

Distance travelled in the last second of its fall is Sn = 24.5m

\(S_{n} = ut + a (n-\frac{1}{2})\\\Rightarrow S_{n} = 0 + g (n-\frac{1}{2})\\\Rightarrow 24.5 = 9.8n – 4.9\\\Rightarrow n = \frac{(24.5 + 4.9)}{9.8}\\\Rightarrow n = 3 seconds.\)

The height of the tower is

\(h = 0 * 0 + 3 + \frac{1}{2} * 9.8(3)^{2}\)

Since s = h, u = 0, a = g = 9.5 m/s

h = 0 + 4.6(3)2

h = 44.1 m

Therefore, the height of the tower, h, is 44.1 m

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