Question

A stone is released from the top of a tower of height 19.6m. Calculate it's final velocity just before touching the ground?

Answer 1

_/\_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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Answer 2

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\pink{ANSWER:-}$

⏩Height of the tower, h= 19.6m.

⏩ Initial velocity, u= 0 m/s.

⏩Final velocity, v=?

Acceleration due to gravity,

g= 9.8 m/s²

→ v²= u² + 2gh

→v²= 0² + 2 × 9.8 × 19.6

→v²= 19.6 × 19.6m

☞ v= 19.6 m/s [The velocity of the stone just before touching the ground]

✔✔✔✔✔✔✔✔

Hope it helps...:-)

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