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Dynamics, falling balls.

by twofish
Tags: balls, dynamics, falling
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twofish
#1
Nov20-08, 05:06 PM
P: 25
1. The problem statement, all variables and given/known data



2. Relevant equations

I'm not sure what equations to use here, or even wether or not I need to use calc. Unfortunately we just started dynamics and most students don't know how to integrate yet (including me)

v = dx/dt
a = dv/dt = d2x/dt2 = v dv/dx


Uniform rectilinear motion: x = x0 + vt
Uniformly accelerated rectilinear motion:
v = vo + at ; x = xo + vot + 1/2at2; v2 = v2o +2a(x-xo)
where o in all the above = initial.


3. The attempt at a solution
I don't even know where to start with this one. hence my question for help.
We know that a = 9.81m/s2 and also that every .5 second there is a ball dropped.
Do i have to integrate anything?
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naresh
#2
Nov20-08, 05:15 PM
P: 166
As you have stated already, a is a constant (9.81 m/s^2). With uniform acceleration, there is no need to know how to integrate.
twofish
#3
Nov20-08, 09:40 PM
P: 25
ok. Well I think I have a solution then, but it's not very fancy (I just used techniques from high school physics).
1) I first contemplate ball #1 to find the final velocity and the time it is at 3m.
Vf2=Vi2+2ax
Vf2= 02+2(9.81)(3)
Vf=7.67m/s
Now the time:
x=( (Vf+Vi)/2)t ; 3m=((7.67+0)/2)t ; t= .78206s

Now I say Ball #2 is dropped .5s after ball #1 and solve for it's final velocity
a=(Vf-Vi)/t ; Vf= 2.767m/s

Now I solve for displacement between ball 1 and 2 according to their respective instantanous(or final) velocities and time being dropped apart from each other.
Where x is displacement.
x=( (Vf+Vi)/2)t
x= ((7.67+2.767)/2)*0.5
x= 2.6097m

Can someone confirm this is correct methodology and answer?
Thanks,

naresh
#4
Nov20-08, 09:52 PM
P: 166
Dynamics, falling balls.

Quote Quote by twofish View Post
ok. Well I think I have a solution then, but it's not very fancy (I just used techniques from high school physics).
High school physics is all I use everyday

1) I first contemplate ball #1 to find the final velocity and the time it is at 3m.
Vf2=Vi2+2ax
Vf2= 02+2(9.81)(3)
Vf=7.67m/s
Now the time:
x=( (Vf+Vi)/2)t ; 3m=((7.67+0)/2)t ; t= .78206s
This part is correct. (It was probably easier to use the formula for displacement, x = Vi t + 1/2 a t2.)

Now I say Ball #2 is dropped .5s after ball #1 and solve for it's final velocity
a=(Vf-Vi)/t ; Vf= 2.767m/s

Now I solve for displacement between ball 1 and 2 according to their respective instantanous(or final) velocities and time being dropped apart from each other.
Where x is displacement.
x=( (Vf+Vi)/2)t
x= ((7.67+2.767)/2)*0.5
x= 2.6097m

Can someone confirm this is correct methodology and answer?
Thanks,
I don't quite understand your explanation for this part of your solution, but if you're saying what I think you're saying, the answer is correct! Are you saying that the problem is equivalent to the distance moved by a ball starting with an initial velocity of 2.767m/s in 0.5 sec?

A simpler solution might be to just find what distance ball #2 travels in 0.28206 seconds.
twofish
#5
Nov20-08, 10:13 PM
P: 25
Actually.. i didn't articulate that very well i suppose. But you're right, if I just solved for the distance ball #2 travelled and deducted that figure from 3m it should tell me the spread between them and hence h (or x).


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