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

Explanation

Initial velocity, `V = ?`
Final velocity, V = ?
Height of tower, `s = 19.6 m`
Acceleration due to gravity,
`g = 9.8 ms^(-2)`
By third law of motion,
`v^(2) - u^(2) = 2gs`
`v^(2) - (0ms^(-1))^(2) = 2 xx 9.8 ms^(-2) xx 19.6 m`
`v^(2) = 19.6 xx 19.6 m^(2)s^(-2)`
`v = sqrt(19.6 xx 19.6 m^(2)s^(-2))`
`v = 19.6 ms^(-1)`