AP EAMCET · PHYSICS · Motion In One Dimension
A ball projected vertically upwards with a velocity 'v' passes through a point \(P\) in its upward journey in a time of ' \(x\) ' seconds. From there, the time in which the ball again passes through the same point \(P\) is
- A \(\frac{\mathrm{v}}{2 \mathrm{~g}}\)
- B \(\frac{2 \mathrm{~V}}{\mathrm{~g}}-\mathrm{x}\)
- C \(\frac{V}{2 g}-x\)
- D \(2\left(\frac{\mathrm{v}}{\mathrm{g}}-\mathrm{x}\right)\)
Answer & Solution
Correct Answer
(D) \(2\left(\frac{\mathrm{v}}{\mathrm{g}}-\mathrm{x}\right)\)
Step-by-step Solution
Detailed explanation
\(t_1 + t_2 = \frac{2v}{g}\) \(x + t_2 = \frac{2v}{g} \Rightarrow t_2 = \frac{2v}{g} - x\) \(t_2 - x = \left(\frac{2v}{g} - x\right) - x = \frac{2v}{g} - 2x = 2\left(\frac{v}{g} - x\right)\)
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