A stone is thrown vertically upward with an initial velocity of 40 m per second.Taking G is equals to 10 M per second square find maximum height reached by stone.What is the net displacement and the total distance covered by the stone.
Answers
According to the equation of motion under gravity:
Where,
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 m s−2
Let h be the maximum height attained by the stone.
Therefore,
0 - (40)2 = 2 x h x (-10)
h= 40 x 40 / 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
_/\_Hello mate__here is your answer--
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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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