A stone is released from the top of tower of height 19.6m. Calculate i

1.	A stone is released from the top of tower of height 19.6m. Calculate it's final velocity. Just before touching the ground.plz don't spam....
Answer» Given:- •Height of the TOWER(h)=19.6m •Acceleration due to gravity(g)=9.8m/s² •Initial velocity(u)=0 (as the stone was at REST INITIALLY) To find:- •Final velocity or velocity of the stone before touching the GROUND(v) Solution:- By using the 3rd equation of motion,when u=0,we get----- =>v²=2gh =>v²=2×9.8×19.6 =>v²=384.16 $= > v = \sqrt{384.16}$ =>v=19.6m/s Thus,velocity of the stone before touching the ground is 19.6m/s.

A stone is released from the top of tower of height 19.6m. Calculate it's final velocity. Just before touching the ground.plz don't spam....

Answer»

Given:-

•Height of the TOWER(h)=19.6m

•Acceleration due to gravity(g)=9.8m/s²

•Initial velocity(u)=0 (as the stone was at REST INITIALLY)

To find:-

•Final velocity or velocity of the stone before touching the GROUND(v)

Solution:-

By using the 3rd equation of motion,when u=0,we get-----

=>v²=2gh

=>v²=2×9.8×19.6

=>v²=384.16

$= > v = \sqrt{384.16}$

=>v=19.6m/s

Thus,velocity of the stone before touching the ground is 19.6m/s.

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A stone is released from the top of tower of height 19.6m. Calculate it's final velocity. Just before touching the ground.plz don't spam....​