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A body is dropped from a height h and a second body is thrown vertically up fromthe ground with a velocity of gh simultaneously. At what height from the ground dothey meet each other? |
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Answer» Answer: A STONE A is dropped from rest from a height h above the ground. A second stone B is simultaneously THROWN VERTICALLY up from a POINT on the ground with velocity v. The line of motion of both the stones is same. Stone A is dropped so initial velocity u = 0 Let's say t time they meet S = ut + 21 at² a = g ( acceleration due to gravity) Distance covered by stone A = 0 + 21 gt² Distance covered = 2h ( mid point) => 2h = 21gt² => h = gt² => t² = GH => t = gh Stone B initial velocity = v a = -g ( as it is going against gravity) 2h = vt - 21gt² putting t = gh => 2h = vgh - 21×ggh => h = vgh => v = hh |
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