Another Mechanics Query (1 Viewer)

[Hermes1]

A solid of unit mass is dropped under gravity from rest at a height of h metres. Air resistance is proportional to the speed v of the mass, acceleration under gravity is g.

okay so a = g-kv

the question is:

using a = v.dv/dx

show that:

the problem is that when v=0, x=h
this means when i solve for the constant, h gets in the way of the answer im supposed to prove.
can i use at x=0, v=0 instead. making the point of projection the origin?, this gives me the answer above, but i dont no if it is legit to do this?

 

[hscishard]

I feel stupid, what is x? the only I got from the question that it's a one dimensional plane
Ok now im thinking x= height, h is a constant

 

[Shadowdude]

x is displacement, correct?

 

[Hermes1]

hscishard, have you done mechanics, bcuz this is a mechanics question. that culd be why u dont understand it

 

[Hermes1]

x is displacement, correct?

x is the displacement. u are quite correct

 

[Drongoski]

Of course you may treat the drop-off point as x=0 if it makes treatment easier.

 

[Hermes1]

Of course you may treat the drop-off point as x=0 if it makes treatment easier.

tnk u very much drongoski, i was worried bcuz i hadnt done anything with h in my answer so thank u very much.

drongoski remind me to buy u lunch after the hsc as a thank you for helping me so much.

 

[Drongoski]

tnk u very much drongoski, i was worried bcuz i hadnt done anything with h in my answer so thank u very much.

drongoski remind me to buy u lunch after the hsc as a thank you for helping me so much.

You are always welcome = since you are polite and acknowledge help given directly to person rendering help.

Not like:" thanks in advance" or "thanks guys" or worse complete silence after helpers have taken quite a bit of trouble to type out solution.

 

[hscishard]

Haven't started mechanics since most of it isn't in my trials but I still can't get the question! Isn't this just 3unit?

 

[Hermes1]

Haven't started mechanics since most of it isn't in my trials but I still can't get the question! Isn't this just 3unit?

this is definitely not 3 unit. this was Q7 from Gosford high school 2010 MX2 trial

3 unit motion is very closely linked to 4 unit motion, you need to understand the basic principles of 3 unit motion to master mechanics

 

[xxxzxc]

yes make x = 0 because it is where you start as you are dropping it from that height

 

[hscishard]

this is definitely not 3 unit. this was Q7 from Gosford high school 2010 MX2 trial

3 unit motion is very closely linked to 4 unit motion, you need to understand the basic principles of 3 unit motion to master mechanics

It really feels like you can solve it using 3unit methods...
a=g-kv

Wouldn't it be a=kv-g? Since gravity is down and resistance is up?

 

[hscishard]

One silly mistake and it stuffs you up lol

I did it with 3U methods using x=0 v=0, I think it's just a hard 3unit motion question

 

[bleakarcher]

if u are considering the downward motion of the particle, when x=0,v=0.
when u are considering the upward motion of the particle, when x=h, v=0

 

[hscishard]

Other way around, unless you want to take account to negative level zero values

...right?

Nvm, I was confused with displacement and x=0 being ground zero or w/e you call it

 

[bleakarcher]

wrong, u take the starting position as the origin alwayz

 

[hscishard]

Then if x = displacement and x=h, then the thing would either be h units up or down the drop off. Right...?
If so, v wouldn't be =0

 

[hscishard]

wrong, u take the starting position as the origin alwayz

What was that directed to? I meant ground zero as to where the thing originally was

 

[bleakarcher]

that is wrong. v would be final velocity, at the instantanteous point in time it hits the ground

 

[hscishard]

Much confused...v would reach final verlocity, that would mean it can't be =0. Say what...?