Question

A ball is released from the top of a tower of height 'h' meter. It takes 'T' seconds to reach the ground. What is the position of the ball at T/3 second?

• A. $\frac{h}{9}m$ from ground
• B. $\frac{7h}{9}$ from the ground
• C. $\frac{8h}{9}$ from ground
• D. $\frac{17h}{9}$ from ground

Answer 1

The correct option is C $\frac{8h}{9}$ from ground

The acceleration of the ball will be g. Initial velocity will be 0. In T seconds, the body travels 'h' meters.
By applying equations of motion we get
$s=ut+\left(1/2\right)g{T}^2$ $\because u=0$
$h=\left(1/2\right)g{T}^2-\left[1\right]$
in $\frac{T}{3}$ sec:
$h_1=\left(1/2\right)g{T}^2=\left(1/2\right)g{\left(\frac{T}{3}\right)}^2=\left(1/2\right)g\left(\frac{T^2}{9}\right)-\left[2\right]$
from
[1] and [2] we get
$h_1=\frac{h}{9}$, which is the distance from the point of release.
Therefore distance from ground is $h-\frac{h}{9}=\frac{8h}{9}$.