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

Explanation

Here, Initial velocity u=0
final velocity v=? (to be calculated )
Acceleration due to gravity `g=9.8 m//s^(2)` (Stone comes down )
And Height h=19.6 m
Now ,we know that for a freely falling body :
`v^(2)=u^(2)+2gh `
So , ` v^(2)=(0)^(2)+2xx9.8xx19.6`
`v^(2)=19.6xx19.6`
`v^(2)=(19.6)^(2)`
v=19.6 m/s
Thus the velocity of stone just before hitting the ground will be 19.6 metres per second .