TS EAMCET · Physics · Motion In One Dimension
A body thrown vertically up to reach its maximum height in \(t\) second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is
- A \(\sqrt{2} t\)
- B \(\left(1+\frac{1}{\sqrt{2}}\right) t\)
- C \(\frac{3 t}{2}\)
- D \(\frac{t}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(B) \(\left(1+\frac{1}{\sqrt{2}}\right) t\)
Step-by-step Solution
Detailed explanation
The ball is thrown vertically upwards then according to equation of motion \[ (0)^2-u^2=-2 g h \] and \[ 0=u-g t \] From Eqs. (i) and (ii), \[ h=\frac{g t^2}{2} \] When the ball is falling downwards after reaching the maximum height…
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