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A stone dropped from the top of a tower is found to travel $\left(\frac{5}{9}\right)$ of the height of the tower during the last second of its fall. The time of fall is ....... $s$
$2$
$3$
$4$
$5$
Solution
(b)
Let the total height of tower $=H$
Total time of journey $=t$
Time taken to cover the $\frac{5 h}{9}$ is = last second
So, $s_t-s_{t-1}=\frac{5 h}{9}$
$\Rightarrow \frac{1}{2} g t^2-\frac{1}{2} g(t-1)^2=\frac{5}{9} \times \frac{1}{2} g t^2$ $\left[\because h=\frac{1}{2} g t^2\right]$
$\Rightarrow \frac{1}{2} g\left(t^2-t^2-1+2 t\right)=\frac{1}{2} g t^2 \times \frac{5}{9}$
$\Rightarrow (2 t-1)=\frac{5}{9} t^2$
$\Rightarrow 18 t-9=5 t^2$
$\Rightarrow 5 t^2-18 t+9=0$
$\Rightarrow 5 t^2-15 t-3 t+9=0$
$\Rightarrow 5 t(t-3)-3(t-3)=0$
$\Rightarrow(5 t-3)(t-3)=0$
$t=\frac{3}{5}, t=3 s \quad\left(t=\frac{3}{5}\right.$, doesn't satisfy the given criterion, so we neglect it)
