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

Answer

The magnitude of the final velocity is `19.6ms^(-1)`.

Explanation

Here, `u=0, g=+9.8ms^(-2), s=+19.6m`
`v^(2)-u^(2)=2gs`
`v^(2)-0=2 xx (+9.8) xx (19.6)`
`v^(2)=19.6 xx 19.6`
`v=pm 19.6ms^(-1)`
Here, `v=+19.6ms^(-1)` because the final velocity is in the downward direction.