Question

Question 14 A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Class 9 - Science - Gravitation Page 144

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

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Answer 2

Answer:

The final velocity of the stone just before touching the ground is 19.6m/s.

Explanation:

Given:

Distance(s)=19.6m

Initial velocity(u)=0m/s

Acceleration(a)=g=9.8m/s²

To find: Final velocity(v)=?

Solution:

By using the second equation of motion,

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

∴$v^{2} =0+2(9.8)19.6$

∴$v^{2}=384.16$

∴$v=\sqrt{384.16}$

∴v=19.6 m/s

Therefore, the final velocity of the stone just before touching the ground is 19.6m/s.