Question

A stone is dropped from the top of a tower. If it travels 34.3m in the last second before it reaches the ground, find the height of the tower? Take g=10m/$s^2$

Answer 1

Step1: Given

Displacement in last second, $S_n=34.3m$

$g=10m{s}^{-2}$

Step 2: Formula used and caculation

For $n^{th}$ second displacement

$S_n=u+\frac{a}{2}(2n-1)$

Here, $S_n=34.3,a=g=10\frac{m}{s^2},u=0$

$\therefore 34.3=0+\frac{10}{2}(2n-1)$

$\Rightarrow 34.3=5(2n-1)$

$\Rightarrow 2n=\frac{34.3}{5}+1$

$\Rightarrow n=4$

Now by using second equation of motion

$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow s=0+\frac{1}{2}\times 10\times 4^2$

$\Rightarrow s=80m$

So, height of the tower is 80 m.