Question

A stone is released from the top of a tower of height 19.6m/sec.calculate its final velocity just before touching the ground.

Answer 1

U = 0
a = 9.8m/sec square
s = 19.6m

Using third equation of motion
v square = u square + 2as

v square = 0 + 2 × 9.8 × 19.6

v square = 384.16
v = √384.16
v = 19.6m/sec

= 19.6m/sec

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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