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

Maximum height =80 m, Net Displacement =0 total distance covered =160 m

Explanation

According to the equation of motion under gravity :
`v^(2) -u^(2) =2 gs`
u= initial velocity of the stone =40 m/s
v= final velocity of the stone =0
s= Height of the stone
g= Acceleration due to gravity `=-10 ms ^(-2)`
Let h be the maximum height attained by the stone therefore ,
`0-(40)2=2xxhxx(-10)`
`h=(40xx40 )/20=80 m`
therefore , total distance covered by the stone during its upward and downward journey `=80+80=160 m`

net displacement of the stone during its upward and downward journey `=80+(-80)=0`