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A stone is dropped from top of a tower 100 m height. At the same instant another stone is thrown vertically upward from the base of the tower with the velocity of 25 m/s. When and where will the two stones meet? Given g = 10 m/s2. |
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Answer» For Stone A, u = 0, S = x and a = + g = 10 m/s2 Formula s = ut + 1/2 at2 Arriving at x = 5t2 —(1) For stone B s = 100 – x, a = – g and u = 25 m/s Arriving at 100 – x = 25t – 5t2 …(2) Obtaining t = 4 second Obtaining x = 80 m |
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