Thread: Mechanical Mathematics

A diver of mass 60 kg, on entering water with an initial velocity of 15.35 ms^-1, experiences a force of resistance, due to the water opposing motion, of magnitude 180v Newtons, where v is the velocity of the diver in ms^-1. In addition to the force of gravity, the diver also experiences a constant upward buoyancy force of 720 N. Take the origin of motion to be at the surface of the water. The magnitude g of the acceleration due to gravity can be taken as 10 ms^-1.

Calculate the maximum depth, in metres, the diver will submerge. Give your answer to 3 decimal places.

The answer is 4.41. Can anyone show me the workings? :'(



Rgds, Bruce

Originally Posted by weijing85

A diver of mass 60 kg, on entering water with an initial velocity of 15.35 ms^-1, experiences a force of resistance, due to the water opposing motion, of magnitude 180v Newtons, where v is the velocity of the diver in ms^-1. In addition to the force of gravity, the diver also experiences a constant upward buoyancy force of 720 N. Take the origin of motion to be at the surface of the water. The magnitude g of the acceleration due to gravity can be taken as 10 ms^-1.

Calculate the maximum depth, in metres, the diver will submerge. Give your answer to 3 decimal places.

The answer is 4.41. Can anyone show me the workings? :'(



Rgds, Bruce

I'm going to set upward to be the positive (vertical) direction.

As always we set

Just about everything is pointed upward save the weight.



From here you can calculate the acceleration. You've got an initial velocity and an acceleration. How do you find the greatest depth from here?

-Dan

Hi dan, can give me some more hints...i still cant get it.

find the acceleration (deceleration) as indicated by -Dan and then apply the Uniform Decelaration formula to find the total space (depth) travelled..

I calculated the acceleration is 5. Am i right?