Question

a stone is released from the top of a tower of height 19.6 M calculate its velocity just before touching the ground

Answer 1

Hey mate,, here is ur ans..

Initial velocity of the stone, u = 0

Final velocity of the stone, v = ?

Height of the tower, h = 19.6 m

Acceleration due to gravity, g = 9.8 ms-2

For a freely falling body:

v² - u² = 2 g h

v²- 0²= 2 × 9.8 × 19.6

v² = 19.6 × 19.6

v = √ (19.6)²

v = 19.6 ms-1

Hence, the velocity of the stone just before touching the ground is 19.6 ms-1.

#hope it helps

Answer 2

$Here \: is \: the \: answer \: of \: your \: question$

`$\textbf{We have to find velocity}$`


According to question we have given,

u (initial velocity) = 0 m/s

s (height) = 19.6 m

a (acceleration) = 9.8 m/s²

Now, formula used here is

v² - u² = 2as

v² = u² + 2as

v² = (0)² + 2 × 9.8 × 19.6

v² = 384.16

v = √384.16

`$\textbf{v = 19.6 m/s}$`

_________________________________

If we take a = 10 m/s²

and u = 0 m/s

s = 19.6 m

Then,

v² = u² + 2as

v² = (0)² + 2 × 10 × 19.6

v² = 392

v = √392

`$\textbf{v = 19.7 m/s}$`