a stone is thrown vertically up with an initial velocity 40 m/s . taking g = 10 m/s ' find the maximum height reached by the stone . what is the net displacement covered by the body . fast please !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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Answered by
1
I think its like this
V2 - U2=2as
0-1600=2*-10*h
H=80m
V2 - U2=2as
0-1600=2*-10*h
H=80m
Answered by
0
Given data ,
u=40m/s
g=10m/s^2
h max =?
h max=u^2/2g
=(40)^2/2×10
=8000m
◆h max=8000m or
converting into km
1km=1000m
1m=1/1000 m
so,8000m=8000×1/1000 km
=8km(h max)
s=?
We know that in vertically projected body s=h
so,s=8km(or)8000m
●◆Thanks for question◆●
u=40m/s
g=10m/s^2
h max =?
h max=u^2/2g
=(40)^2/2×10
=8000m
◆h max=8000m or
converting into km
1km=1000m
1m=1/1000 m
so,8000m=8000×1/1000 km
=8km(h max)
s=?
We know that in vertically projected body s=h
so,s=8km(or)8000m
●◆Thanks for question◆●
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