Question

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

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

Answer 1

The correct option is A $\frac{8h}{9}$ meters from the ground

We know that $s=ut+\frac{1}{2}g{t}^2$,
or $h=\frac{1}{2}g{T}^2(\because u=0)$
now for $\frac{T}{3}$ second, vertical distance moved is given by
$h^{\prime }=\frac{1}{2}g{\left(\frac{T}{3}\right)}^2\Rightarrow h^{\prime }=\frac{1}{2}\times \frac{g{T}^2}{9}=\frac{h}{9}$
$\therefore$ position of ball from ground $=h-\frac{h}{9}=\frac{8h}{9}$