If a freely falling body travels in the last second a distance equal to the distance travelled by it in the fi

Question

If a freely falling body travels in the last second a distance equal to the distance travelled by it in the first three seconds, the time of the travel is:

Accepted Answer

1. **Analyze the Motion:** The body starts from rest ($u=0$) and undergoes free fall with acceleration $g$. Let the total time of flight be $t$ seconds. The 'last second' corresponds to the $t$-th second.
2. **Distance in First 3 Seconds:** Using the equation $s = ut + \frac{1}{2}at^2$:
   $$ S_3 = 0 + \frac{1}{2}g(3)^2 = \frac{9g}{2} $$ .
3. **Distance in Last ($t$-th) Second:** The distance covered in the $n$-th second is given by $D_n = u + \frac{a}{2}(2n - 1)$. Here $n=t$:
   $$ D_t = 0 + \frac{g}{2}(2t - 1) $$
4. **Equate and Solve:** The problem states $D_t = S_3$.
   $$ \frac{g}{2}(2t - 1) = \frac{9g}{2} $$
   $$ 2t - 1 = 9 $$
   $$ 2t = 10 \implies t = 5 	ext{ s} $$

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