Question

heya

A stone is released from top of tower of height 19.6m .Caculate the final velocity just before touching the ground

plz answer this..​

Answer 1

$\huge{\underline{\underline{.........Answer}}}$

S = H = 19.6metres

U = 0 m/s

A =G = 9.2 ^2

$\huge{\underline{\underline{.....Explanation}}}$

As H,U,G are given so we should 3rd newton equation

$\huge{\boxed{\boxed{ v^2 - u^2 = 2as }}}$

so,substitute the values

v^2 - 0^2 =2 × 9.8 × 19.6

v = $\sqrt{ 19.6× 19.6}$

v = 19.6 m/s

$\huge{\boxed{\boxed{v = 19.6 m/s}}}$

so,the velocity before touching the ground is

19.6 m/s

Answer 2

S = H = 19.6metres

U = 0 m/s

A =G = 9.2 ^2

As H,U,G are given so we should 3rd newton equation

v^2- u^2= 2as

so,substitute the values

v^2 - 0^2 =2 × 9.8 × 19.6

v =

v = 19.6 m/s

so,the velocity before touching the ground is

19.6 m/s