Question

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

Answer 1

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

MARK AS BRAINLIEST

Answer 2

Answer:

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