A stone is thrown vertically upwards with an initial velocity and is caught back. By taking g=10m/s find the maximum height reached by the stone. What is the displacement and the total distance covered by the stone.
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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`.
Explanation:
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