Question

A particle is dropped from the top of the tower. During its motion it covers 9/25 th part of the tower in the last one sec. Find the height of the tower.

Answer 1

Given,

The height of tower $=H$

Total time is taken to reach the ground $=T$

$\frac{16H}{25}=0.5g{T}^2——-\left(1\right)$

$\frac{9H}{25}=0.5g(T-1{)}^2$

Divide above two equations

$\frac{16}{9}=(\frac{T}{(T-1)}{)}^2$

$\frac{4}{3}=\frac{T}{(T-1)}$

$3T=4T-4$

$T=4s$

Put the value of$T=4s$ in equation (1)

$\frac{16H}{25}=0.5g\times 4^2$

$H=\left(\frac{25}{16}\right)0.5\times 9.8\times 16$

$H=122.5m$

THus, Height of the tower is $122.5m$