A stone is thrown verticaly upwards with a velocity of 40m/s and is caught back.Takinf g =
10m/s², calculate the maximum height reached by the stone.What is the net displacement and total distance covered by the stone?
Answers
Answered by
1
Explanation:
maximum height =u^2/2g
=40^2/2*10=1600/20=80m.
displacement =0m.
total distance =v^2/2a-u^2/2a
=0^2/2*10-40^2/2*10
=0-1600/20
=-80m.
atif463:
1600/20=80
oh ya
my mistake
it's all right
in my answer sheet ,the total distance = 160 m
So u^2 must have to be 3200
Answered by
1
Answer:
Explanation:
We have take equation 2as=V^2-u^2
V=final velocity (0)
U=initial velocity (40)
2×10×s=0^2-40^2
20s=-1600
So,s=1600/20
=80meters upwards
its wrong because 40^2 is not equal to 3600
Now I correct it answer is 80
no problem
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