Question

A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in $T/3$ seconds?

• A. $h/9$ metres from the ground
• B. $7h/9m$ from the ground
• C. $8h/9$ metres from the ground
• D. $17h/18m$ from the ground

Answer 1

The correct option is C $8h/9$ metres from the ground

the acceleration of the ball will be g. Initial velocity will be 0.

in T sec. body travels h mts.

by applying equations of motion we get

$s=ut+\left(1/2\right)g{T}^2$

$h=\left(1/2\right)g{T}^2$ ------[1]

in $T/3$ sec

$h_1=\left(1/2\right)g{T}^2=\left(1/2\right)g(\frac{T}{3}{)}^2=(1/2\left)g\right(\frac{T^2}{9})$ -------[2]

from

$\left[1\right]$ and $\left[2\right]$ we get $h_1=\frac{h}{9}$ distance from point of release.

therefore distance from ground is $h-\frac{h}{9}=\frac{8h}{9}$