Question

A stone is released from the top of a towerof height 19.6m. Calculate its final velocity just before touching the ground ​

Answer 1

$\bf\large\underline\pink{Answer:-}$

U= 0

a = 9.8 m/sec square

s = 19.6 m

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 = 19.6 m/sec

ANSWER = 19.6 m/sec

HOPE IT HELPS :):):):):)

Answer 2

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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Explanation: