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4. A ball is released from top of the tower of hight hm. It takes t sec. to reach the ground. What is the position of ball inT/3 sec.

mohit , 11 Years ago
Grade 8
anser 1 Answers
Ashish Kumar

Last Activity: 11 Years ago

h=1/2 gt^2 , t=sq.root( 2h/g)

in t/3 secs the height through which the ball will fall , H=1/2 gt^2/9= h/9 m
so position from the ground = h-h/9 = 8h/9 m



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