Question

A ball is released from the top of a tower of height $h$. It takes time $T$ to reach the ground. What is the position of the ball (from ground) after time $T/3$?

• A. $h/9$ m
• B. $7h/9$ m
• C. $8h/9$ m
• D. $17h/18$ m

Answer 1

The correct option is C $8h/9$ m

Given,

Let the acceleration be ${}^{\prime }{g}^{\prime }$

Then initial velocity$=0m/s$ [in T seconds]

Let the distance be ${}^{\prime }{h}^{\prime }m.$

now,

Equation of motion.

$s=ut+\frac{1}{2}g{t}^2$

We have $h=\frac{1}{2}g{T}^2$
In $T/3$ second, distance fallen = $\frac{1}{2}g{\left(\frac{T}{3}\right)}^2=\frac{h}{9}$
So position of the ball from the ground is
$h-\frac{h}{9}=\frac{8h}{9}$m