• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Need some physics help

F

Finns14

Golden Member
I have a number of questions but I guess I will go one at a time here is the first.

A basketball player makes a jump shot. The 0.569-kg ball is released at a height of 2.27 m above the floor with a speed of 8.05 m/s. The ball goes through the net 3.20 m above the floor at a speed of 4.79 m/s. What is the work done on the ball by air resistance, a nonconservative force?

I guess I have to consider both the x and y direction but I am still rather lost. Any help would be greatly appreciated. Also for you jerks who are going to say do your own homework I am asking for explantions not answers. Thanks again.
 
The easiest way to do it would be to use conservation of energy. Calculate how much energy (potential + kinetic) the ball had when it was released, and again when it went through the basket. The difference in energy is how much it lost due to air resistance, and thus the work done.
 
I can give the answer if it helps anyone because the site that I do homework on generates new values for the question
 
Originally posted by: Heisenberg
The easiest way to do it would be to use conservation of energy. Calculate how much energy (potential + kinetic) the ball had when it was released, and again when it went through the basket. The difference in energy is how much it lost due to air resistance, and thus the work done.
Click to expand...



Ok this is making some sense to me let me try and see if I can get it
 
Originally posted by: Heisenberg
The easiest way to do it would be to use conservation of energy. Calculate how much energy (potential + kinetic) the ball had when it was released, and again when it went through the basket. The difference in energy is how much it lost due to air resistance, and thus the work done.
Click to expand...

Agreed, this is the first method that I would have suggested
 
Originally posted by: PurdueRy
Originally posted by: Heisenberg
The easiest way to do it would be to use conservation of energy. Calculate how much energy (potential + kinetic) the ball had when it was released, and again when it went through the basket. The difference in energy is how much it lost due to air resistance, and thus the work done.
Click to expand...

Agreed, this is the first method that I would have suggested
Click to expand...



I tried this method and I can't see to get it I'm using the formula 1/2mv^2+mgh calculate that for both half parts and subtract the difference and i can't seem to get it
 
Originally posted by: Finns14
I have a number of questions but I guess I will go one at a time here is the first.

A basketball player makes a jump shot. The 0.569-kg ball is released at a height of 2.27 m above the floor with a speed of 8.05 m/s. The ball goes through the net 3.20 m above the floor at a speed of 4.79 m/s. What is the work done on the ball by air resistance, a nonconservative force?

I guess I have to consider both the x and y direction but I am still rather lost. Any help would be greatly appreciated. Also for you jerks who are going to say do your own homework I am asking for explantions not answers. Thanks again.
Click to expand...

conservation of energy.

work = change in kinetic + change in potential

you have all teh variables. just plug and chug.
 
Originally posted by: Finns14
I have a number of questions but I guess I will go one at a time here is the first.

A basketball player makes a jump shot. The 0.569-kg ball is released at a height of 2.27 m above the floor with a speed of 8.05 m/s. The ball goes through the net 3.20 m above the floor at a speed of 4.79 m/s. What is the work done on the ball by air resistance, a nonconservative force?

I guess I have to consider both the x and y direction but I am still rather lost. Any help would be greatly appreciated. Also for you jerks who are going to say do your own homework I am asking for explantions not answers. Thanks again.
Click to expand...

m*g*h1 + 1/2*m*(v1)^2 = m*g*h2 + 1/2*m*(v2)^2 + F

Plug and chug and solve for F
 
Originally posted by: PurdueRy
Originally posted by: Finns14
I have a number of questions but I guess I will go one at a time here is the first.

A basketball player makes a jump shot. The 0.569-kg ball is released at a height of 2.27 m above the floor with a speed of 8.05 m/s. The ball goes through the net 3.20 m above the floor at a speed of 4.79 m/s. What is the work done on the ball by air resistance, a nonconservative force?

I guess I have to consider both the x and y direction but I am still rather lost. Any help would be greatly appreciated. Also for you jerks who are going to say do your own homework I am asking for explantions not answers. Thanks again.
Click to expand...

m*g*h1 + 1/2*m*(v1)^2 = m*g*h2 + 1/2*m*(v2)^2 + F

Plug and chug and solve for F
Click to expand...


Guess it must be and arithmatic error
 
OK, I predict you will spend most of the week staying home, watching TV, and on Saturday night, you will see a hot girl on the Internet, causing you to touch yourself.

EDIT: Oh, I'm sorry! I thought you needed some PSYCHIC help. Misread that. Sorry.
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top