Question

A stone is thrown vertically upward with an initial velocity of 40 m s. Find tha maximum height reached by the stone. What is the displacement and the total distance covered by the stone? (g =10 m /s²)

Answer 1

Use the 3rd equation of Motion v square =u square +2as                             v=0 u=40 g= -10 find s

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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