Question

a stone is thrown vertically upwards with a velocity of 40m/s and is caught back calculate the maxiimum height reached by the stone what is the net displacement and total distance covered by the stone

Answer 1

where V = 0

U = 40 m/s

a = -g m/s² = -10 m/s²

0 = (40)² -2 × 10 × S

S = 80 m .

hence maximum hight is reached by stone = 80 m .

now, stone comes back to intial point.

hence

total distance covered by stone = 80 ( in upward motion ) + 80( in downward motion ) = 160 m .

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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