A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.

To determine the final velocity of a stone released from the top of a tower, falling freely under the influence of gravity, we can apply principles of kinematics and the laws of motion.

Given:
– Initial velocity (u) = 0 m/s (as the stone is released from rest)
– Height of the tower (s) = 19.6 meters
– Acceleration due to gravity (g) = 9.81 m/s² (approximately)

Using the equations of motion under constant acceleration, the one relating initial velocity (u), final velocity (v), acceleration (a), and displacement (s) is:

v² = u² + 2as

Here, (v) represents the final velocity, (u) is the initial velocity, (a) is the acceleration due to gravity, and (s) is the displacement (height).

Given that the initial velocity (u) is 0 m/s, the equation simplifies to:

v² = 0 + 2 x 9.81 x 19.6
v² = 0 + 2 x 9.81 x 19.6
v² = 0 + 2 x 192.276
v² = 384.552

Taking the square root of both sides to find (v):

v = √(384.552)
v ≈ 19.6 m/s

Hence, the final velocity of the stone just before touching the ground is approximately 19.6 m/s. This calculation assumes the absence of air resistance or other external forces affecting the stone during its fall.