Question

A ball is dropped from the top of a tower during the last second of its free fall it covers a distance of 55 metre what is the height of the tower ?(g=10m/s)​

Answer 1

Given,

=> Distance transversed in last second = u+ g/2 ( 2n-1)

$S_{nth} = u + \frac{g}{2}(2n-1)$

$55 = 0 + \frac{10}{2}(2n-1)$

$55 = 5(2n-1)$

$55 = 10n-5$

$55 + 5 = 10n$

$60 = 10n$

$n = \frac{60}{10}$

$n = 6 s$

Again,

Height of Tower,

$=> h= ut+ \frac{1}{2}g{t}^{2}$

$=> h = 0 + \frac{1}{2}\times10\times6\times6$

$=> h = 5\times6\times6$

$\bold{=> h = 180 m}$

Hence, the height of Tower = h = 180 m