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From the foot of a tower 90m high stone is thrown up so as to reach the top of a tower two seconds later another stone is dropped from height of Tower with two stones meet at |
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Answer» of the building (d) = 90mLet’s find the velocity of STONE thrown up:We know, v^2 – u^2= 2adVelocity of stone after reaching top of the building = 0Acceleration is opposite to acceleration due to gravity = -9.8m/s^20- u^2= 2*(-9.8)*90- u^2=-1764Therefore, velocity of stone MOVING up= 42m/sDistance at time t = h(t) =ut-1/2at^2h(t) = 42t-4.9t^2 -----(1)h1(T) = HEIGHT of building – 1/2aT^2T=t-2 (since second stone is thrown after two seconds)H1= 90 – 4.9(t-2)^2When the two stones meet, distance are same. So, Set h= h142t - 4.9t^2 = 90 -4.9t^2 +19.6t -19.642t-19.6t=90-19.622.4t = 70.4 => t = 3.14sSubstitute the VALUE of “t” in equation (1) h =83.6m |
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