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2) r8 m(3) 15.6 m(4) 39.2 mTwo bodies are thrown vertically úpward, with the same initial velocity of 98 m/s but 4 sec apartHow long after the first one is thrown when they meet ?(2) 11 sec(1) 10 sec(3) 12 sec(4) 13 sec |
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Answer» When they meet, their displacements will be same. Therefore , s= ut-(1/2)gt^2=u(t-4) -(1/2)g(t-4)^2 Simplifying and making t subject, we get t=(u+2g)/g. OR Substituting the values, t=(98+19.6)/9.8 OR t=12s after the first body was thrown. Go and ask this question to your teacher 👌 |
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