Kinematics Question 243
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 [AIEEE 2004]
Options:
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
Show Answer
Answer:
Correct Answer: C
Solution:
$ \because \ \ h=ut+\frac{1}{2}gt^{2} $
$ \Rightarrow h=\frac{1}{2}gT^{2} $ After $ \frac{T}{3} $ seconds, the position of ball, $ {h}’=0+\frac{1}{2}g{{( \frac{T}{3} )}^{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. $
