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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