Answer to Question #135753 in Physics for Aum Patel

Question

5. A porpoise jumps straight up and crashes back into the water at 8.9 m/s. The drag and buoyancy forces of the water slow the porpoise down with an acceleration of -9.3 m/s2 as the porpoise finally slows to a stop. Find the depth the porpoise reaches.

Answer 1

Determine the initial conditions (data) that would help to solve the problem.

Initial velocity right before hitting the surface of water: 8.9 m/s

Motion occurs in water stopping the porpoise

The porpoise experiences deceleration of 9.3 m/s²

The porpoise stops after some distance h

Let's get it solved!

All we need is a simple expression that ties initial velocity, deceleration, and distance together:

$$h=\frac{v^2}{2g}=\frac{8.9^2}{2\cdot9.3}=4.3\text{ m}.$$