Question

The cliff divers at acapulco, mexico, jump off a cliff 17.2 m above the ocean. ignoring air resistance, how fast are the divers going when they hit the water?

Answer 1

The divers are in free fall, and they cover a distance of
`S=17.2 m` in a uniformly accelerated motion, with an acceleration equal to
`g=9.81 m/s^2` (gravitational acceleration). We can find their final velocity vf just before hitting the water using the following relationship

`2aS = v_f^2 -v_i^2`
where
`v_i=0` is their initial velocity, which is zero because the divers start from rest. Therefore, substituting the numbers we can find vf:

`v_f = √(2aS)= √(2\cdot 9.81 m/s^2 \cdot 17.2 m)=18.4 m/s`