If the distance travelled by a freely falling body in the last second of its journey is equal to the distance travelled in the first 2s, the time of descent of the body is

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1.5s

2.5s

answer is C.

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Detailed Solution

We know

$\begin{array}{l} S_{n} = \frac{1}{2} g (2 n - 1) & S_{2} = \frac{1}{2} g (2^{2}) \\ ∴ \frac{1}{2} g (2 n - 1) = \frac{1}{2} g \times 4; 2 n - 1 = 4 \Rightarrow 2 n = 5 \\ \Rightarrow n = \frac{5}{2} = 2.5 \end{array}$

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