Question

a stone released from top of the tower of height 19.6 cm .calculate its final velocity just before touching the ground( take g=9.8m/s^2)

Answer 1

V^2 -u^2=2gh

V^2=2×9.8×19.6(initial velocity is zero)

V^2=384

V=√384

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______❤

✿┅═══❁✿ Be Brainly✿❁═══┅✿