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. $



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