AP EAMCET · PHYSICS · Motion In One Dimension
A body is projected vertically upwards with a velocity \(u\) from the top of a tower. Time taker by it to reach the ground is \(n\) times, then the time taken by it to reach the highest point in its path. Height of the tower is
- A \(\frac{n u^2(n-1)}{2 g}\)
- B \(\frac{n u^2(n-2)}{g}\)
- C \(\frac{n u^2(n-2)}{2 g}\)
- D \(\frac{u^2}{2 g}(n+1)\)
Answer & Solution
Correct Answer
(C) \(\frac{n u^2(n-2)}{2 g}\)
Step-by-step Solution
Detailed explanation
Let the time taken to reach the maximum height, when thrown vertically upwards \[ t_1=\frac{u}{g} \] If \(t_2\) be the time to hit the ground, then given \[ t_2=n t_1=n u / g \]…
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