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.

Q&A -Ask Doubts and Get Answers

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

Using the third equation of motion,

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

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