A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds - Physics MCQ AIEEE 2004 2026 | ExamTest.live

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A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T /3 seconds?

Solution & Step-by-step Explanation

For the full fall from height h in time T : h = \frac{1}{2} g T^{2} \dots (i) Distance traveled from the top in time t = T /3 is: h^{'} = \frac{1}{2} g (\frac{T}{3})^{2} = \frac{1}{9} (\frac{1}{2} g T^{2}) Using equation (i) : h^{'} = \frac{h}{9} This is the distance from the top. The position from the ground is: H_{g ro u n d} = h - h^{'} = h - \frac{h}{9} = \frac{8 h}{9}

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A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T /3 seconds?

h /9 metres from the ground

7 h /9 metres from the ground

8 h /9 metres from the ground

17 h /18 metres from the ground

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