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 \(=10 \mathrm{~ms}^{-2}\) )
- A 9
- B 36
- C 25
- D 49
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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