Question

if an object is dropped from the top of the tower the distance it covered in the last second is 7 times the distance it covers in the first second find the time of flight

Answer 1

Distance covered by a freely falling object in the $n^{th}$ second is given as:

$S_n=u+\frac{\text{g}}{2} (2n-1)\\ \\ \text{where\ u = initial velocity}\\ \\\text{g = acceleration due to gravity}$

Given that the distance it covered in the last second is 7 times the distance it covered in the first second.

Let the last second is the $n^{th}$ second.

$S_{n}=7 \times S_{1}\\ \\\Rightarrow u+\frac{\text{g}}{2} (2n-1)=7 \times [u+\frac{\text{g}}{2} (2 \times 1-1)]\\ \\\Rightarrow 0+\frac{\text{g}}{2} (2n-1)=7 \times [0+\frac{\text{g}}{2} (2-1)]\\ \\\Rightarrow \frac{\text{g}}{2} (2n-1)=7 \times [\frac{\text{g}}{2}]$

$\\ \\\text{cancelling } \frac{\text{g}}{2}\ \text{from both sides}\\ \\\Rightarrow 2n-1=7\\ \\ \Rightarrow 2n=7+1=8\\ \\\Rightarrow n = \frac{8}{2}\\ \\\Rightarrow n = 4\ seconds$

Thus, total time is 4 seconds.

Answer 2

4 sec is the right answer using the Gravitational Laws