Question

A ball is released from the top of a tower of height $h$ m. It takes $T$ s to reach the ground. What is the position of the ball in $\frac{T}{3}$ s?

• A. $\frac{h}{9}$ m from the ground
• B. $\frac{7h}{9}$ m from the ground
• C. $\frac{8h}{9}$ m from the ground
• D. $\frac{17h}{18}$ m from the ground

Answer 1

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

Lets consider the equation $S=ut+\frac{1}{2}a{t}^2$
Here $u=0,a=g$
So $S=\frac{1}{2}g{t}^2$
Now $h=\frac{1}{2}g{T}^2$
The distance traveled by ball in $\frac{T}{3}$ s will be $S=\frac{1}{2}g\times {\left(\frac{T}{3}\right)}^2=\frac{h}{9}$
So height of ball above ground = (Total Height) $-$ (Distance Traveled in $\frac{T}{3}$ s)
$=h-\frac{h}{9}=\frac{8h}{9}$ m above ground