Question

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

• A. $100m$
• B. $125m$
• C. $175m$
• D. $120m$

Answer 1

The correct option is B $125m$

Let the height of the tower $=H$

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

So, $H=0.5g{T}^2........\left(1\right)$

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

Dividing both of the above equation.

$\frac{25}{16}=\frac{T^2}{(T-1{)}^2}$

$\Rightarrow \frac{5}{4}=\frac{T}{T-1}$

$\Rightarrow T=5s$

Put this value in equation 1 then we get,

$H=0.5g\times 5^2\Rightarrow H=125m$