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Question

A ball is released from the top of a tower of height $$h$$. It takes time $$T$$ to reach the ground. What is the position of the ball (from ground) after time $$T/3$$?


A
h/9 m
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B
7h/9 m
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C
8h/9 m
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D
17h/18 m
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Solution

The correct option is C $$8h/9$$ m
Given,
Let the acceleration be $$'g'$$
Then initial velocity$$ = 0 m/s$$ [in T seconds]
Let the distance be $$'h' m.$$
now,
Equation of motion.
$$s=ut+\dfrac{1}{2}gt^{2}$$
We have $$h=\dfrac{1}{2}gT^2$$
In $$T/3$$ second, distance fallen = $$\dfrac{1}{2}g{ \left( \dfrac { T }{ 3 }  \right)  }^{ 2 }=\dfrac { h }{ 9 } $$
So position of the ball from the ground is 
$$h-\dfrac{h}{9}=\dfrac{8h}{9}$$m

Physics

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