A stone is thrown vertically upward with an initial velocity of `40 ms^(-1)`. Taking `g= 10 ms^(-2).` find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone ?

Answer

The maximum height reached is 80m.
The total distance covered is 160m. Net displacement is zero.

Explanation

Here, `u=+40ms^(-1), g=-10ms^(-2)`,
The final velocity, v is the maximum height is zero.
`v^(2)-u^(2)=2gs`
`0-(40)^(2)=2 xx (-10)xxs`
`s=(40 xx 40)/(2 xx 10)=80m`
The total distance covered =80m+80m
=160m
Net displacement =80m-80m=0