[ChaseD13]

Alright... AP physics suckssss....

I'd appreciate any help.

Here's the problem (on motion in a resistive medium)

"An object of mass m is thrown vertically upward with a speed V(0). Assuming that air resistance is proportional to the instantaneous velocity and that the constant of proportionality is k, show that the time to reach maximum height is given by the following.

Tmax= (m/k) ln [1 + (kv0/mg0)] "

m= mass
g=gravity

The point of this problem is to use implicit differentiation to go from acceleration to velocity, and then to set velocity to 0 because that's what the velocity would be at max height.

Any help?

 

[myinitialsaredac]

Wouldn't a derivative from acceleration yield rate of change of acceleration or the snap?

I believe an integral from acceleration yields velocity.

Sorry I haven't done implicit differentiation in some time so other than that I am no help.

Cheers,
Dave

 

[ChaseD13]

oops my bad, i kind of wrote the wrong thing- youtake the anti derivative of acceleration to find velocity, and the antiderivative of velocity to find position. ahhhh its so hard!

basically, i need to fill in the work that gets to that answer, but for the life of me i can't get it

 

[myinitialsaredac]

It may be of some help or it may not that the (m/k) in the front is a constant and also that I believe you need to do a u substitution.
d/dx ln (u) = u'/u
so T=(m/k)ln(u)du
u= 1+(kv0/mg0) du= this is where you need to do the top times the derivative of the bottom minus the bottom times the derivative of the top all over the bottom scared and ultimately your implicit differentiation.
Then you should end up with T'=(m/k)(u'/u)
Then set T'=0 and solve that should give you the extrema
Good Luck,
Dave

 

[myinitialsaredac]

I believe they gave you the position function and you need to do a first derivative test to find the extrema and apply it to the original equation.

Right?

Dave

 

[ChaseD13]

hmm, im looking at it now, and im not sure what im given. If it makes sense, i believe i have the position function where velocity is 0. Right? Should i just take the derivative twice to show acceleration?

 

[myinitialsaredac]

I also believe you have the position function, but I believe it is enough to take the first derivative and show where velocity is 0. That is where the object is not moving and changing from moving upward to moving downward. If you show where acceleration is 0 it is where the object is beginning to decelerate, or accelerate in the opposite direction, slowing the velocity to zero.

Dave

 

[ChaseD13]

hmm... thanks. I'll see what i can come up with. Too bad its 1130 and school starts at 655 haha...

Thanks again I appreciate it!

- physics is the bane of my existence

 

[myinitialsaredac]

Very welcome, Sorry I wasnt more help.

I had a calculus problem the other day that was a pain in the backside hah.

Good luck man!

Dave

 

[kingkevo25]

Man I feel dumb. I'm a college physics major well into quantum mechanics and I'm not even sure what the problem is asking....

 

[Borat]

 

[myinitialsaredac]

Quote:

Originally Posted by Borat /img/forum/go_quote.gif

LOL

That is hilarious!!!!

Dave