question_answer

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                                [AIEEE 2004]

A)                         h/9 meters from the ground                 

B)                         7h/9 meters from the ground            

C)                         8h/9 meters from the ground   

D)                         17h/18 meters from the ground

Correct Answer: C

Solution :

                \[\because \ \ h=ut+\frac{1}{2}g{{t}^{2}}\]\[\Rightarrow h=\frac{1}{2}g{{T}^{2}}\]                         After \[\frac{T}{3}\] seconds, the position of ball,             \[{h}'=0+\frac{1}{2}g{{\left( \frac{T}{3} \right)}^{2}}=\frac{1}{2}\times \frac{g}{9}\times {{T}^{2}}\]             \[h'=\frac{1}{2}\times \frac{g}{9}\times {{T}^{2}}\]\[=\frac{h}{9}m\] from top             \ Position of ball from ground \[=h-\frac{h}{9}=\frac{8\ h}{9}m.\]