A stone is thrown vertically upwards with an initial velocity of 40 m/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.

Explanation

According to the eqution of motion under gravity, `v^(2)-u^(2)=2gs`
Where v=final velocity of the stone=o
u=Initial velocity of the stone =40m/s
s=Height of the stone
g=Acceleration due to gravity `=-10ms^(2)` Let h be the maximum height attained by the stone.
Therefore, `o-(40)^(2)=2xxhx(-10)`
`h=40xx40//20=80m`
Therefore, total distance covered by the stone during its upward and downward journey `=80+80=160m`
Net displacement of the stone during its upwards and downwards journey.
`=80+(-80)=0`