Question

If stone is released from the top of a lower of weight 19.6m. Calculate its final velocity just before touching the ground?

Answer 1

Answer:

V = 19.6 m/s

Explanation:

Use Newton's law of motion:

${v}^{2} + {u}^{2} = 2as$

Here, the initial velocity is zero and acceleration is 'g' and distance is height of the tower:

${v}^{2 } = 2gh$

Now substitute g = 9.8 and h = 19.6

${v}^{2} = 2 \times 9.8 \times 19.6$

${v}^{2} = 384.16$

Hence,

$v = 19.6 \: mps$

Hope that you will find this useful :)

Mark me 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