Question

A stone is thrown vertically upward with a velocity of 40 m/s taking g=10m/s .Find the maximum height reached by the stone .What is net distance covered by the stone

Answer 1

${v}^{2} - {u}^{2} = 2gh$
${0}^{2} - {40}^{2} = 20h$
$1600 \div 20 = h$
$h = 80meters$
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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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