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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

  1. A \(\frac{n u^2(n-1)}{2 g}\)
  2. B \(\frac{n u^2(n-2)}{g}\)
  3. C \(\frac{n u^2(n-2)}{2 g}\)
  4. D \(\frac{u^2}{2 g}(n+1)\)
Verified Solution

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 \[ t_2=n t_1=n u / g \]…