InterviewSolution
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InterviewSolution
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A stone is thrown verticaly upward with an initial velocity of `40m//s`. Taking `g=10m//s^(2)`, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone? |
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Answer» Here, `u=40m//s, g=-10m//s,h=?,v=0` From `v^(2)-u^(2)=2gh, 0-(40)^(2)=2(-10)h or h=(40xx40)/20=80m` As final position of the stone coincides with its initial position, net displacement =0. Total distance coverd by the stone `=h+h=2h=2xx80m=160m`. |
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