Question

A ball is released from the top of a tower of height h metre . it takes T second to reach the ground . what is the position of the ball in T/3 second ? [AIEEE 2004] ​

Answer 1

Explanation :- second law of motion gives

• S = ut + 1/2gT²
• or , h = 0 + 1/gT² [∵ u = 0]
• T = √(2h/g)
• At t = T/3s , s = 0 + 1/2g(T/3)²
• or, s = 1/2g•(T²/9)
• => s = g/18 ×2h/g
• s = h/9 m
• Hence, the position of ball from the ground
• => h - h/9 = 8h/9 m Answer

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