Question

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

Answer 1

s = ut + 0.5at^2

19.6 = 0 + 0.5 x 9.81 x t^2

19.6 = 0.5 x 9.81t^2

39.2 = 9.81t^2

t^2 = 3.99

t = 1.99 s

mgh = 0.5 mv^2

2gh = v^2

2 x 9.81 x 19.6 = v^2

v^2 = 384.552

v = 19.6 m/s

Acceleration of Free-falling object is always 9.81 m/s^2

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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