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.​

Answer 1

Initial Velocity u=0

Fianl velocity v=?

Height, s=19.6m

By third equation of motion

v

2

=u

2

+2gs

v

2

=0+2×9.8×19.6

v

2

=384.16

⇒v=19.6m/s

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Answer 2

Answer:

here is your answer

Explanation:

According to the equation of motion under gravity:

v2 − u2 = 2 gs

Where,

u = Initial velocity of the stone = 0

v = Final velocity of the stone

s = Height of the stone = 19.6 m

g = Acceleration due to gravity = 9.8 m s-2

∴ v2 − 02 = 2 × 9.8 × 19.6

v2 = 2 × 9.8 × 19.6 = (19.6)2

v = 19.6 m s-1

Hence, the velocity of the stone just before touching the ground is 19.6 m s-1.