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.

Answer 1

Answer:

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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Explanation:

Answer 2

Initial Velocity (u) = 0m/s

Final Velocity (v) = unknown

Height dropped from (h) = 19.6 m

Acceleration = Acceleration due to Gravity = 10 m/s^2

Using the relation:

v^2 = u^2 + 2as

v^2 =(0)^2 + 2(10)(19.6)

v^2 = 392

v = √392

v = 19.8 m/s

Hence, the final velocity achieved just before hitting the ground is v = 19.8m/s