Question

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?

Answer 1

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`.