AP EAMCET · PHYSICS · Motion In One Dimension
A ball is dropped from a tower of height \(80 \mathrm{~m}\). The time it takes to cover the last \(50 \%\) of its fall is
(acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A \(2 \sqrt{2} \mathrm{~s}\)
- B \(1.17 \mathrm{~s}\)
- C \(4 \mathrm{~s}\)
- D \(2.0 \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(1.17 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
The given situation is shown below For Ist part of journey, \(\begin{aligned} & u=0, a=-g=-10 \mathrm{~ms}^{-2} \\ & s=-40 \mathrm{~m}\end{aligned}\) Now using, \(v^2-u^2=2 a s\) we get, \(\quad v^2-0=2 \times(-10)(-40)\) \(\Rightarrow \quad v^2=800\)...(i) And using…
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