Question

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)​

Answer 1

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

Answer 2

_/\_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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