Who is released from the top of the tower of height 19.6 m calculate its final velocity just before touching the ground?

Sol. Given,

Initial velocity, u = 0

Final velocity, v =?

Acceleration due to gravity, g = 9.8 m/s2

Height, h = 19.6 m

We know that,

v2 = u2 + 2gh

 = (0)2 + 2 × 9.8 × 19.6

 = 19.6 × 19.6

 = (19.6)2

 = 19.6 m/s

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

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Question:

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:

Given:

u = Initial velocity of the stone = 0

v = Final velocity of the stone = ?

s = Height of the tower = 19.6 m

g = Acceleration due to gravity = 9.8

Who is released from the top of the tower of height 19.6 m calculate its final velocity just before touching the ground?

Using the third equation of motion,

Who is released from the top of the tower of height 19.6 m calculate its final velocity just before touching the ground?

Who is released from the top of the tower of height 19.6 m calculate its final velocity just before touching the ground?

Who is released from the top of the tower of height 19.6 m calculate its final velocity just before touching the ground?

Who is released from the top of the tower of height 19.6 m calculate its final velocity just before touching the ground?

Hence, the velocity of the stone just before touching the ground is 19.6

Who is released from the top of the tower of height 19.6 m calculate its final velocity just before touching the ground?
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Who is released from the top of the tower of height 19.6 m calculate its final velocity just before touching the ground?

Who is released from the top of the tower of height 19.6 m calculate its final velocity just before touching the ground?

Who is released from the top of the tower of height 19.6 m calculate its final velocity just before touching the ground?