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 is the net displacement and the total distnce covered by the stone?

Explanation

Here `u=40ms^(-1)`
`a=-g=-10ms^(-2)`
v=0
s=?
Using 2 as `v^(2)-u^(2)` we get
`2(-10)S=0^(2)-(40)^(2)`
or `S=(40xx40)/20` m
`=80m`
`:.` Maximum height reached by the stone =80 m
Net displacement of the stone =80-80=0
Total distance covered by the store `=80+80`
`=160m`