Question

a stone is released from a height of 5 m .calculate its final velocity just before touching the ground ? ( take g = 10 m per second square )

Answer 1

1. v=(2xgxh)^1/2 so (2x5x10)^1/2=10m/s

Answer 2

_/\_Hello mate__here is your answer--

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u = 0 m/s

v = ?

s = Height of the stone = 5 m

g = 10 ms−2

According to the equation of motion under gravity

v^2 − u^2 = 2gs

⇒ v^2 − 0^2 = 2 × 10 × 5

⇒ v^2 = 100

⇒ v = 10 ms−1

Hence, the velocity of the stone just before touching the ground is 10 ms^−1.

I hope, this will help you.☺

Thank you______❤

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