mechanics q (1 Viewer)

[john-doe]

a body of mass m is projected vertically upwards with speed u, air resistance is k times the square of the speed, where k is a constant. find the speed of the body when it is next at the point of projection.

i got the answer as -u. But i think thats only possible where air resistance is neglected so i might have made a mistake..Help! thanks

 

[john-doe]

bump

 

[barbernator]

do you want a hint or solution?

 

[asianese]

You have to find when it comes back down as well.

 

[barbernator]

Ok here is the gist of it. when the particle is travelling upwards, its equation of acceleration is a=-g-kv^2. So integrate dx/dv to find x in terms of v. When you have evaluated the constant, find the x value when v=0. This x value will be used for our next acceleration constant.

Now that the particle has reached its peak, the equation if motion changes as resistance is in opposition to gravity. so a=g-kv^2. integrate again dx/dv to find x in terms of v. Find the constant by using the point (x=0,v=0 (we are resetting the start for new equation)) and then substitute in x=(displacement you had found before) and solve for v.

 

[john-doe]

do you want a hint or solution?

anything?

You have to find when it comes back down as well.

alright i got the following expression for x in terms of v...

x=(m/2k).ln(ku^2+mg)-(m/2k)ln(kv^2+mg)...now they want the velocity at point of projection so i let x=0..

i end up with v^2=u^2 therefore v= -u ....where did i go wrong??

an

 

[john-doe]

its equation of acceleration is a=-g-kv^2.

i got a=-g-kv^2/m ???

 

[barbernator]

anything?



alright i got the following expression for x in terms of v...

x=(m/2k).ln(ku^2+mg)-(m/2k)ln(kv^2+mg)...now they want the velocity at point of projection so i let x=0..

i end up with v^2=u^2 therefore v= -u ....where did i go wrong??

an

you have to realise that there are different equations for motion as the particle is going up and down, so you have to solve the first, and then use this to evaluate the second. Also, for ease, you don't need mass in your equations as k is just a simple constant rather than k/m or whatever as they mean the same thing, especially as this question doesn't require an evaluation of the constant.

 

[john-doe]

you have to realise that there are different equations for motion as the particle is going up and down, so you have to solve the first, and then use this to evaluate the second. Also, for ease, you don't need mass in your equations as k is just a simple constant rather than k/m or whatever as they mean the same thing, especially as this question doesn't require an evaluation of the constant.

can you just upload your working out if u dont mind?? i wanna see it! thanks if u can

 

[D94]

Ok here is the gist of it. when the particle is travelling upwards, its equation of acceleration is a=-g-kv^2. So integrate dx/dv to find x in terms of v. When you have evaluated the constant, find the x value when v=0. This x value will be used for our next acceleration constant.

Since resistance is a force, F_resistance = kv². Then, a = (±)g-kv²/m. Yes/no?

 

[barbernator]

Since resistance is a force, F_resistance = kv². Then, a = (±)g-kv²/m. Yes/no?

but because the constant is the same for both equations, this doesn't matter right?

 

[D94]

but because the constant is the same for both equations, this doesn't matter right?

Quite possibly, I rushed through my solution, so I will have a look again.

 

[barbernator]

Quite possibly, I rushed through my solution, so I will have a look again.

ill do a solution without it and see if its the same

 

[barbernator]

my answer. <a href="http://www.codecogs.com/eqnedit.php?latex=v=\sqrt{\frac{u^2g}{g-ku^2}}~where~k~includes~mass" target="_blank"><img src="http://latex.codecogs.com/gif.latex?v=\sqrt{\frac{u^2g}{g-ku^2}}~where~k~includes~mass" title="v=\sqrt{\frac{u^2g}{g-ku^2}}~where~k~includes~mass" /></a>
can someone confirm?

 

[john-doe]

my answer. <a href="http://www.codecogs.com/eqnedit.php?latex=v=\sqrt{\frac{u^2g}{g-ku^2}}~where~k~includes~mass" target="_blank"><img src="http://latex.codecogs.com/gif.latex?v=\sqrt{\frac{u^2g}{g-ku^2}}~where~k~includes~mass" title="v=\sqrt{\frac{u^2g}{g-ku^2}}~where~k~includes~mass" /></a>
can someone confirm?

the answer for the velocity at the point of projection is

u^2/ {sqrt [1+ (ku^2/mg)]} note the k in this answer doesnt include mass

 

[barbernator]

the answer for the velocity at the point of projection is

u^2/ {sqrt [1+ (ku^2/mg)]} note the k in this answer doesnt include mass

shit ok ill fix tomorrow and post solution. too tired

 

[john-doe]

shit ok ill fix tomorrow and post solution. too tired

lol i have this question in assessment tomorrow morning...but thanks for trying..ill ask my mate and hope it doesnt show up in the test! hahaaha

 

[D94]

the answer for the velocity at the point of projection is

u^2/ {sqrt [1+ (ku^2/mg)]} note the k in this answer doesnt include mass

Yes, that's what I got. I used F_resistance = kv² for my calculations. Initial equation was: ma = mg + kv² and second equation was: ma = -mg + kv²

Edit: wait... was the answer with a big square root sign?

 

[john-doe]

Yes, that's what I got. I used F_resistance = kv² for my calculations. Initial equation was: ma = mg + kv² and second equation was: ma = -mg + kv²

care to share the workin out???

 

[Aysce]

Yes, that's what I got. I used F_resistance = kv² for my calculations. Initial equation was: ma = mg + kv² and second equation was: ma = -mg + kv²

Edit: wait... was the answer with a big square root sign?

Wouldn't the initial equation where the particle is vertically projected be ma = -mg-kv^2 ?