InterviewSolution
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InterviewSolution
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A ball thrown up vertically returns to the thrower after 8 second. Calculate (i) velocity with which it was thrown (ii) maximum height it acquired. (iii) velocity with which it hit the ground.Given `g=9.8 m//s^(2)`. |
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Answer» As time of ascent =time of descent`=(1)/(2)` total time. `:. t=(8)/(2)=4s` (i)`u=?,v=0,a=-g9.8m//s^(2)`. From `v=u+at` `0=u-9.8xx4` or `u=39.2m//s` (ii)From `v^(2)-u^(2)=2a s` `0-(39.2)^(2)=2(-9.8)h` `h=(-(39.2)^(2))/(-2xx9.98)=78.4m` (iii) velocity with which body hits the ground = velocity of projection of body `=39.2m//s`. |
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