Projectile motion along one dimension -- urgent help needed! (1 Viewer)

[blackops23]

Sorry but I have no idea what to do with this question:

Q. A stone is let fall from a tall building and 1 second later another stone is projected vertically downwards with a velocity of 20m/s. When will the second stone overtake the first?



Can someone please explain to me what I am meant to do, and why?

Thanks, appreciate the help

 

[zxreth]

interesting, haven't encountered a question like this before, got no idea how to do it though. wait for drongoski

 

[Liability]

interesting, haven't encountered a question like this before, got no idea how to do it though. wait for drongoski

that question is piss easy, why is a 4unit student asking this?

Derive the two displacement equations for the ball (the first one starts one second before the second, that's not to compensate for) and equate. lol newb

Don't you do 4unit mechanics, that's a lot harder than this.

I just don't get you 4unit students, you can do the harder questions but fail on the easy ones.

 

[taeyang]

that question is piss easy, why is a 4unit student asking this?

Derive the two displacement equations for the ball (the first one starts one second before the second, that's not to compensate for) and equate. lol newb

Don't you do 4unit mechanics, that's a lot harder than this.

I just don't get you 4unit students, you can do the harder questions but fail on the easy ones.

Wow man, I never thought of that, and I'm sure OP and zxreth didn't either.

 

[Liability]

blackops, stop asking newb questions.

 

[Liability]

Wow man, I never thought of that, and I'm sure OP and zxreth didn't either.

lol u sarcastic fucker, the little 3unit kiddo that thought he was so smart on the other question from blackops.

Are you sure you have the brains to do engineering?

 

[taeyang]

Oh shit, that's right, you got 98 in 2 unit, didn't you? Sorry.

 

[blackops23]

that question is piss easy, why is a 4unit student asking this?

Derive the two displacement equations for the ball (the first one starts one second before the second, that's not to compensate for) and equate. lol newb

Don't you do 4unit mechanics, that's a lot harder than this.

I just don't get you 4unit students, you can do the harder questions but fail on the easy ones.

soz but yeah I'm pretty noob at maths (i just started projectile/SHM a few days ago):

Ok so deriving the displacement equation for the first stone:

x'' = g = 9.8 (I made the downwards direction be positive)
x' = gt + c --> x' = 0 when t = 0, therefore x' = gt

x = g(t^2)/2 + D ---> displacement = 0 at t=0,

therefore equation (1) ----> x = g(t^2)/2

--------------

Now what do I do for equation 2, I mean how do I compensate for the "1 second later" business?

 

[bleakarcher]

you can not determine the time at which the second particle will overtake the first as the initial heights both particles are not given. you could possibly assume they are launched from the same height however and obtain then a numerical answer

 

[Reserves]

you can not determine the time at which the second particle will overtake the first as the initial heights both particles are not given. you could possibly assume they are launched from the same height however and obtain then a numerical answer

ur an idiot, you can. That's why it says "launched from a VERY tall building"

All you need to know is where the first stone is at time t=1 , then derive the two eqns of displacement and equate. NEWBS

When t=1 the first stone is at x= g/ 2

 

[pwoh]

a = g
v = gt
x = (1/2)gt^2

A = g
V = g(t-1) + 20
X = (1/2)g(t-1)^2 + 20(t-1)

X = x

...

t = (20-1/2 g) / (2 - g)

[Check tho, not sure if right]

 

[blackops23]

X = (1/2)g(t-1)^2 + 20(t-1)

Can someone please verify if this is right? If it is, then t = 1.48 is the answer, meaning that the second stone would only take 0.48 seconds to overtake the first stone... i don't know if that is a plausible answer...

 

[Provisions]

I can confirm that you are a dumb fuck lololol.

fucking 14.5/15 for microeconomics midsem and 28/32 for accounting midsem. To think, I was only considering tutoring maths, I am a fucking genius at everything commerce.

Going to drop the engineering side of my degree and concentrate on commerce subs, actuary major, put my maths genius to good use etc etc.

My current WAM is 75 , by the end of this semester I reckon it will be like 80 and who knows where it will be in a yrs time.

HAHAH should have got tutoring fail!

The benifits of selfishness, serves you guys XD, you cunts only go to places with high fees and high ATARS and high WAMS, seem to like veerything high.

 

[blackops23]

X = (1/2)g(t-1)^2 + 20(t-1)

Can someone please verify if this is right? If it is, then t = 1.48 is the answer, meaning that the second stone would only take 0.48 seconds to overtake the first stone... i don't know if that is a plausible answer...

 

[cutemouse]

Going to drop the engineering side of my degree and concentrate on commerce subs, actuary major, put my maths genius to good use etc etc.

Not having a go at you or anything, but you know actuary is really difficult right? If you can't do engineering then you'll probably find actuary maths really difficult as well... Just my $0.02.

And don't even get me started on the fact that doing actuary in university is really only doing 40% of the qualifications actually become an actuary (ie. to earn big $$)....

 

[username30]

Not having a go at you or anything, but you know actuary is really difficult right? If you can't do engineering then you'll probably find actuary maths really difficult as well... Just my $0.02.

And don't even get me started on the fact that doing actuary in university is really only doing 40% of the qualifications actually become an actuary (ie. to earn big $$)....

I am not dropping engineering because the maths is hard you newb (in fact I want to change to actuary because I want to more maths!, see how people make these "mathematical models" and see how useful and accurate they really are). I just really don't like the computing, the structure of the labs and I really cannot see myself doing an engineering job (even if I did finish the engineering side of the degree). I will just end up working in the business area anyway and I don't think i'll wake up on day with an urge to change to an engineering job (and even if I do, I will have forgotten everything about engineering by then).

People said 4unit hsc maths was difficult and that made me make the wrong choice and drop it. I want to try actuary and see how overrated it really is.

 

[cutemouse]

I just really don't like the computing

Well, in ACTL2001, I believe you learn and must be able to code the R programming language... I don't know much about other subjects but I wouldn't be surprised if there were more computing requirements.

 

[username30]

Well, in ACTL2001, I believe you learn and must be able to code the R programming language... I don't know much about other subjects but I wouldn't be surprised if there were more computing requirements.

That doesn't seem too bad. Wow..learn a stats computer package and learn excel as well, nothing hard about that. It looks like all you really need to know are assignments of variables and loops. I am good with all that so it shouldn't be a problem. Just graphs kinda pissed me off and actual data structures. Well at least if I transfer now I will have hardly wasted any of my 1.5yrs of engineering subjects.

MATH1141/MATH1241 exempts me from MATH1151/MATH1251
COMP1911/COMP1921 should help with the actuarial programming.

The other courses I did will just count as my general education courses and i'm set.

 

[OldMathsGuy]

Sorry but I have no idea what to do with this question:

Q. A stone is let fall from a tall building and 1 second later another stone is projected vertically downwards with a velocity of 20m/s. When will the second stone overtake the first?



Can someone please explain to me what I am meant to do, and why?

Thanks, appreciate the help

The trick to this is making the maths simple by choosing the right frames of reference.
For this make t=0 when the 2nd stone is thrown, and make the height of the building be y=0.

For the second stone, y=(-gt^2)/2-20t
For the first stone, at t=0, y'=-g and y=(-g)/2
and so y=(-gt^2)/2-gt-g/2

Solving simultaneously, you get t=g/[2(20-g]

At g=10, t=0.5 seconds
(and y=-11.25m)

If you take g=9.8 then these values are slightly different again.

Best Regards
OldMathsGuy

 

[hscishard]

t1 = time for first stone
t2 = time for second stone
t2 = t1 -1
Just use t-1 for the second projectile equation

I got 1.48 seconds after the first rock is thrown