Question

A particle is dropped from top of tower During its motion its covers $\frac{9}{25}$ part of height of tower in last $1sec$ . then height of tower?

• A. $100m$
• B. $125m$
• C. $175m$
• D. $123m$

Answer 1

Answer: (D)

$123m$

Lets consider, $T=$total time taken to reach the ground

$H=$ height of the tower

Initial velocity of the particle is zero.

$g=9.8m/s^2$

According to the question,

From the 2nd equation of motion,

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

$\frac{16H}{25}=\frac{1}{2}g{T}^2$. . . . . . . . .(2)

Divide above two equation, we get

$\frac{16}{9}=[\frac{T}{T-1}{]}^2$

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

$3T=4T-4$

$T=4sec$

Substitute the value of $T$ in equation (2),

$\frac{16H}{25}=\frac{1}{2}\times 9.8\times 4\times 4$

$H=\frac{25}{16}\times 78.4=123m$

The correct option is D.