Again stones?

I stand the top of a building \(80\text{m}\) high and my friend Abhay stands at the ground floor of this building. Then, I drop a stone and Abhay throws a stone vertically upward simultaneously. If it is known that Abhay threw the stone at a velocity of \(20\text{m/s}\), then after how much time (in seconds) would the two stones meet?

Details and assumptions:-

  • The acceleration due to gravity(\(\text{g}\)) is \(10{\text{m/s}}^2\).
  • The motion of the stones are in a straight vertical line perpendicular to the surface.
  • The friction of air is negligible.
  • Give your answer to two decimal places.

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