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

Solution

## The correct option is C $$8h/9$$ mGiven,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 $$h-\dfrac{h}{9}=\dfrac{8h}{9}$$mPhysics

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