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TS EAMCET · Physics · Motion In One Dimension

A body is falling freely from the top of a tower of height 125 m . The distance covered by the body during the last second of its motion is \(x \%\) of the height of the tower. Then \(x\) is (Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )

  1. A 9
  2. B 36
  3. C 25
  4. D 49
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Answer & Solution

Correct Answer

(B) 36

Step-by-step Solution

Detailed explanation

Here \(\mathrm{u}=0\) So \(\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^2\) or, \(125=0+\frac{1}{2} \times 10 \times \mathrm{t}^2\) \(\therefore \mathrm{t}=5 \mathrm{~s}\) The distance covered by the body during the last second of its motion is \(\mathrm{x} \%\) of the height…
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