A stone is thrown vertically upward with an initial velocity of $40m/s$. Taking $g=10m/s^2$, find the maximum height reached by the stone. What's the net displacement and the total distance covered by the stone?

According to the equation of the motion under gravity

$v^2-u^2=2gs$

$u=$initial velocity of the stone$=40m/s$

$v=$ Final velocity of the stone$=0m/s$

Let h be the maximum height attained by the stone

Therefore,

$0^2-{40}^2=2(-10)h$

$h=(40\times 40)/20=80$

Therefore, total distance covered by the stone during its upward and downward journey$=80+80=160m$

Net displacement during its upward and downward journey$=80+(-80)=0$