Question

A stone is released from the top of a tower of height 19.6m.what will be its final velocity just below touching the ground

Answer 1

S=19.6 M, U=0,g=9.8
we know,
2as=v^2-u^2
2gs=v^2-u^2
2*9.8*19.6=v^2
19.6*19.6=v^2
v=19.6

Answer 2

_/\_Hello mate__here is your answer--

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u = 0 m/s

v = ?

s = Height of the stone = 19.6 m

g = 9.8 ms−2

According to the equation of motion under gravity

v^2 − u^2 = 2gs

⇒ v^2 − 0^2 = 2 × 9.8 × 19.6

⇒ v^2 = 2 × 9.8 × 19.6 = (19.6)^2

⇒ v = 19.6 ms−1

Hence, the velocity of the stone just before touching the ground is 19.6 ms^−1.

I hope, this will help you.☺

Thank you______❤

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