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

Answer

19.6 m/s

Explanation

According to the equation of motion under gravity :
` v^(2)-u^(2)=2gs`
where,
u- initial velocity of the stone =0
v= Final velocity of the stone
s= Height of the stone `=19.6 m`
g= Acceleration due to gravity `=9.8 ms^_(-2)`
` therefore v^(2)-0^(2)=2xx9.8 xx19.6xx19.6`
`v^(2)=2xx9.8xx19.6 (19.6)^(2)`
` v=19.6 ms^(-1)`
Hence , the velcoity of the stone just before toching the ground is 19.6 `ms^(-1)`