Question

A stone is released of a tower of height 19.6m . Calculate its final velocity just before touching the growing

Answer 1

initial velocity =0 u=0
equation of motion of 3rd equation
v^2=u^2+2as
v^2= 2×10×19.6
v=root 392
v=19.79

Answer 2

_/\_Hello mate__here is your answer--

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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