The Speed of Cliff Divers at Acapulco, Mexico

Calculating the Final Velocity of Cliff Divers

Cliff divers at Acapulco, Mexico jump off a cliff that is 17.2 meters above the ocean surface. In this scenario, we will ignore the effects of air resistance and calculate how fast the divers are going when they hit the water.

The divers are in free fall, meaning they are only influenced by gravity. They cover a distance of S = 17.2 meters in a uniformly accelerated motion, with an acceleration equal to g = 9.81 m/s^2 (gravitational acceleration).

To find their final velocity vf just before hitting the water, we can use the following relationship:

\[ 2aS = v_f^2 - v_i^2 \]

Where vi = 0 is their initial velocity, which is zero as the divers start from rest. Substituting the given values, we can calculate vf:

\[ v_f = \sqrt{2aS} = \sqrt{2 \cdot 9.81 m/s^2 \cdot 17.2 m} = 18.4 m/s \]

Therefore, the speed of the cliff divers when they hit the water is 18.4 m/s.

What is the final velocity of the cliff divers at Acapulco, Mexico when they hit the water?

The final velocity of the cliff divers at Acapulco, Mexico when they hit the water is 18.4 m/s.