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A stone falls frecly from rest and the total distance covered by it in the last second of its motion cquals the distance covered by it in the first three seconds of its motion. The stone remains in the air for |
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Answer» 6S From s = ut`+(1)/(2) "GT"^(2)` Here `U =0 :. S =(1)/(2) "gt"^(2)` Total distance travelled by the stone inlast second is `D= s_(t) s_(t-1)=(1)/(2) "gt"^(2)-(1)/(2) g (t-1)^(2)= (g)/(2) (2t-1)` Distance travelled by the stone in first three seconds is `s_(3) =(1)/(2) xxgxx3^(2)=(9)/(2) g or 2t-1 =9` t=5s |
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