Using kinematics to find motions with constant acceleration

[Crusaderking1]

Homework Statement

painter is standing on scaffolding that is raised at constant speed. As he travels upward, he accidentally nudges a paint can off the scaffolding and it falls a distance 16.0 m to the ground. You are watching, and measure with your stopwatch that it takes a time of 3.26 s for the can to reach the ground. Ignore air resistance.

What is the speed of the can just before it hits the ground?

Another painter is standing on a ledge, with his hands a distance 4.35 m above the can when it falls off. He has lightning-fast reflexes and if the can passes in front of him, he can catch it. Does he get the chance?

Homework Equations

v=v_o+at

y-y₀=v₀t+.5at²

The Attempt at a Solution

I received 11.1 m/s using the second equation and then plugged that number into the first one for an answer of -20.8 m/s.

Is this right? Thanks!

I am still working on the second question.

 

[Rayquesto]

your initial velocity will be the constant speed being lifted up which you don't know yet.

Its constantly being affected by gravity.

so, Vf=V0 - gt

and Vf^2=V0^2 + 2gy

but velocity just before it hit the ground at the instant is dy/dt=16meters/3.26seconds

It doesn't seem as though you'd need to find V0, but you might for the second part of the question. Let me check though. Hold on. God damn. I don't understand the second part of the question. if he is on a ledge but also above the can, how can he also catch it if the can will always be below?

 

[Crusaderking1]

your initial velocity will be the constant speed being lifted up which you don't know yet.

Its constantly being affected by gravity.

so, Vf=V0 - gt

and Vf^2=V0^2 + 2gy

but velocity just before it hit the ground at the instant is dy/dt=16meters/3.26seconds

It doesn't seem as though you'd need to find V0, but you might for the second part of the question. Let me check though. Hold on. God damn. I don't understand the second part of the question. if he is on a ledge but also above the can, how can he also catch it if the can will always be below?

Oh, I did the first part wrong then?

I have a solution for the second problem, but if my first part is wrong, then it is wrong.

Ya, I don't really get the 2nd question either.

 

[Rayquesto]

If I were you, I'd just say it is impossible, because he'd have to be below which is impossible.

 

[Crusaderking1]

If I were you, I'd just say it is impossible, because he'd have to be below which is impossible.

As much as I would like to agree(based on common sense?), I received an answer that he can, but only based on if -20.8 m/s is correct.

 

[Rayquesto]

show me exactly how you got 11.1m/s.
according to my calculation, 11.1m/s is actually the initial velocity; the constant velocity the scaffolding rose.

 

[Crusaderking1]

show me exactly how you got 11.1m/s.
according to my calculation, 11.1m/s is actually the initial velocity; the constant velocity the scaffolding rose.

Ya, I then use the initial velocity to find for v(t) which using the first equation I obtained -20.8 m.

Then, for b, the answer should be above 4.35 meters I think, so I just use v²=v_o²+2a(y_f-y_o)
Which equals 6.28 meters.

 

[Rayquesto]

oh ****! excuse my language i think youre right, but...sorry let me think for a few minutes on this.

 

[Crusaderking1]

oh ****! excuse my language i think youre right, but...sorry let me think for a few minutes on this.

haha no problem. I definitely appreciate the help!

 

[Rayquesto]

I think the velocity at the instant is actually vy(3.26)=1-1.1 + -19.81(3.26)
Im really sorry about the first answer I gave you that was completely incorrect. I thought it was right. I have to figure out a way to prove how incorrect it is.

 

[Rayquesto]

forgive me its not 1-1.1 its -11.1m/s

 

[Crusaderking1]

forgive me its not 1-1.1 its -11.1m/s

nevermind

 

[Rayquesto]

oh wait then 11.1 is actually positive. sorry man I am screwing up today. Let me show you more. hold on.

 

[Crusaderking1]

ok haha. =)

 

[Rayquesto]

ok so I have a big brain fart right now. You'll really have to excuse me. This comes to me pretty fast but this question is worded strangely. they said it falls a distance of 16 meters, but it also rises from the point that it will start falling, which suggests that the 16/3.26m/s might actually be right.

if you want to know how much time it took to fall 16 meters, you'll probably have to assume that since it rises then goes to a velocity of 0m/s, you can consider that your initial velocity from the time its 0m/s to the time it goes down. So, for that times it takes

16=1/2gt^2

t=1.8 seconds. which means that the paint thing rose for 3.26-1.8 seconds.

knowing this is pretty important because then 0=v0-gt when velocity is zero

so, V0=14.3m/s
god I might be overanalyzing this, but it seems right now.

But I still am very sceptical about the velocity
im pretty sure now that its just gah idk i hope I helped though. what do you think?

 

[Crusaderking1]

ok so I have a big brain fart right now. You'll really have to excuse me. This comes to me pretty fast but this question is worded strangely. they said it falls a distance of 16 meters, but it also rises from the point that it will start falling, which suggests that the 16/3.26m/s might actually be right.

if you want to know how much time it took to fall 16 meters, you'll probably have to assume that since it rises then goes to a velocity of 0m/s, you can consider that your initial velocity from the time its 0m/s to the time it goes down. So, for that times it takes

16=1/2gt^2

t=1.8 seconds. which means that the paint thing rose for 3.26-1.8 seconds.

knowing this is pretty important because then 0=v0-gt when velocity is zero

so, V0=14.3m/s
god I might be overanalyzing this, but it seems right now.

But I still am very sceptical about the velocity
im pretty sure now that its just gah idk i hope I helped though. what do you think?

I found this site with a problem very similar to this one. I don't think the can rose but I don't really know. In this similar situation, I don't think it was accounted for if it did.

http://www.baskent.edu.tr/~aerdamar/ch2ex.pdf Problem 2.89

I checked my work using this site and I think -20.8 is right, although I'm still not sure. What do you think? Thanks.

Does part B look good as well?

 

[Rayquesto]

so, if the actual time it took to get down from rest is 1.8 seconds, then Vy(1.8)=-1/2g(1.8)^2= -15.89m/s

 

[Crusaderking1]

so, if the actual time it took to get down from rest is 1.8 seconds, then Vy(1.8)=-1/2g(1.8)^2= -15.89m/s

nevermind. thanks

 

[Rayquesto]

the only way you can account for the 3.26 seconds automatically is if its in free fall or maybe if its going some initial velocity upwards. ok ok ok yea you are righ

heres the proof.

y=v0t - 1/2gt^2

16=v03.26 - 1/2g 3.26^2

then v0=20.898

so, vy(3.26s)=20.898 - 9.81(3.26s)=-11.1m/s
whew yes you are right

 

[Crusaderking1]

the only way you can account for the 3.26 seconds automatically is if its in free fall or maybe if its going some initial velocity upwards. ok ok ok yea you are righ

heres the proof.

y=v0t - 1/2gt^2

16=v03.26 - 1/2g 3.26^2

then v0=20.898

so, vy(3.26s)=20.898 - 9.81(3.26s)=-11.1m/s
whew yes you are right

Thanks for your time! Does part B look good as well?
I think I understand the constant acceleration equations a little better after this discussion.

 

[Rayquesto]

i am sure its -11.1m/s but why was my earlier post deleted?

 

[Rayquesto]

oh nevermin dit was just on the second page,but for the second part of the question, it just doesn't seem logical with the givens to prove that it is not possibl even though it really isnt.

 

[Crusaderking1]

oh ok. Thanks. I will now stand on a ledge and make it my life goal to catch it.