TS EAMCET · Physics · Motion In One Dimension
A body is falling freely from a point \(A\) at a certain height from the ground and passes through points \(B, C\) and \(D\) (vertically as shown below) so that \(B C=C D\). The time taken by the particle to move from \(B\) to \(C\) is \(2 \mathrm{~s}\) and from \(C\) to \(D\) is 1 s. Time taken to move from \(A\) to \(B\) in seconds is
- A 0.6
- B 0.5
- C 0.2
- D 0.4
Answer & Solution
Correct Answer
(B) 0.5
Step-by-step Solution
Detailed explanation
Let velocity of the particle at point \(B\) be \(v\). Now, \(\quad B C=2 v+\frac{1}{2} g \times(2)^2\) or, \[ B C=2 v+2 g \] Similarly, \[ 2 B C=3 v+\frac{1}{2} \times g \times(3)^2 \] \[ \Rightarrow \quad 2 B C=3 v+\frac{9 g}{2} \] From Eq. (i),…
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