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 in $T/3$ seconds?

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

Answer 1

Answer: (C)

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

$\Rightarrow h=\frac{1}{2}g{T}^2$



After $\frac{T}{3}$ seconds, the position of the ball,

$h^{\prime }=0+\frac{1}{2}g(\frac{T}{3}{)}^2$

$=\frac{1}{2}\times \frac{g}{9}\times T^2$

$h^{\prime }=\frac{1}{2}\times \frac{g}{9}\times T^2=\frac{h}{9}m$ from top

$\therefore$ Position of ball from ground

$=h-\frac{h}{9}=\frac{8h}{9}m$.