Question

A body is dropped from certain height H. If the ratio of the distances travelled by it in (n-3) seconds to (n-3)${}^{th}$ second is 4 : 3, find H. (Take g $=$ 10 m s${}^{-2}$)

Answer 1

$S_{n-3}=ut+\frac{1}{2}a{t}^2$

($u=0,a=g$)

$S_{n-3}=\frac{1}{2}g{t}^2=\frac{1}{2}g(n-3{)}^2$

$S_n^{th}=u+\frac{a}{g}(2n-1)$ [distance traveled in $n^{th}$ second with constant acceleration]

$S_{(n-3)}^{th}=\frac{g}{2}\left(2\right(n-3)-1)$

$=\frac{g}{2}(2n-7)$

$\frac{S_{n-3}}{S_{n-3}^{th}}=\frac{g(n-3{)}^2}{2\times \frac{g}{2}\times (2n-7)}=\frac{4}{3}$

$3(n^2+9-6n)=4(2n-7)$

$3n^2+27-18n=8n-28$

$3n^2-26n+55=0$

$3n^2-15n-11n+55=0$

$3n(n-5)-11(n-50=0$

$n=5,\frac{11}{3}$

$H=\frac{1}{2}g(5{)}^2$

$H=125m$