a stone is thrown vertically upward with an initial velocity of 40 m per second find the maximum height reached by the stone what is the net displacement and total distance covered by the stone( take G equal to 10 M per second square)
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Answered by
6
hey the answer is ....
u = 40m/s v= 0 a = 10
DISTANCE - v^2 - u ^2 = 2as
0 - 40^2 = 2 × 10 × s
1600/20 = s
s = 80m
u = 40m/s v= 0 a = 10
DISTANCE - v^2 - u ^2 = 2as
0 - 40^2 = 2 × 10 × s
1600/20 = s
s = 80m
Answered by
2
_/\_Hello mate__here is your answer--
____________________
u = 40 m/s
v = 0 m/s
s = Height of the stone
g = −10 ms^−2 ( upward direction)
Let h be the maximum height attained by the stone.
Using equation of motion under gravity
v^2 − u^2 = 2gs
⇒0^2 − 40^2 = 2(−10)ℎ
⇒ ℎ =40×40/ 20 = 80 m
Therefore, total distance covered by the stone during its upward and downward journey
= 80 + 80 = 160 m
Net displacement during its upward and downward journey
= 80 + (−80) = 0
I hope, this will help you.☺
Thank you______❤
_______________________❤
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