Question

a stone is released from the top of a tower of height 19. 6 M calculate its final velocity just before touching the ground. please give short answer​

Answer 1

By applying equation of motion -:v2=u2+2gh.

as gravity is always constant =9.8m/s and u (initial velocity)=0.

v2=0+2×9.8×19.6

v2=384.16

v=19.6m/s2

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______❤

__________________________❤