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22 Jan 09
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Originally posted by joe shmoplot the position function
The entire thing until between step 4 and 5 is un-necessary
If you pay attention, step 5 is the exact equation I derived.
let me ask you this, why do you feel it is necessary to find the max height. It is irrelevant to this problem.?
P(t) = -4.9t^2 + 11.5cos(49.5)t + 1.6
It is a graph of the path of the flight, as you will see the object velocity changes from positive to negative..only the zero of this equation is when the ball falls 1.6 m further than the designated initial position of zero(ie y=-1.6)
Im looking at the exact form of this problem in a physics text right now
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22 Jan 09
2 edits
Originally posted by joe shmoalso go while you are at it, take the derivative of P(t) and set it equal to 0
plot the position function
P(t) = -4.9t^2 + 11.5sin(49.5)t + 1.6
It is a graph of the path of the flight, as you will see the object velocity changes from positive to negative..only the zero of this equation is when the ball falls 1.6 m further than the designated initial position of zero(ie y=-1.6)
Im looking at the exact form of this problem in a physics text right now
P'(t) = -9.8t + 11.5sin(49.5) = 0
t= -11.5sin45.9/-9.8
t = 0.8426....sec..I bet that is what you have found the value of time to be from you first equation that pairs with the maximum hieght
what makes you think that the Position function P(t) suddenly stops working..It is a concave downward parabola that continues indefinately
then if you want to plot in your velocity vectors as tangent lines
here are the equations
y = 8.257t + 1.6 ( the equation of the line tangent to the point (t,P(t)) at t =0
and for your time t = 1.6852.. ( when you decide to use the position equation)
y = -8.257t + 15.514
and while your at it go ahead and plot the line
y=1.6
And i can assure you that all of the equations were all derived from
dt/dx[P(t)] the function you use in step 5
your way isnt wrong
but it is the long way around
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22 Jan 09
Originally posted by joe shmosorry, that notation above is supposed to be
also go while you are at it, take the derivative of P(t) and set it equal to 0
P'(t) = -9.8t + 11.5sin(49.5) = 0
t= -11.5sin45.9/-9.8
t = 0.8426....sec..I bet that is what you have found the value of time to be from you first equation that pairs with the maximum hieght
what makes you think that the Position function P(t) suddenly stops working.. ...[text shortened]... dx[P(t)] the function you use in step 5
your way isnt wrong
but it is the long way around
dP/dt[P(t)] or the derivative of position with respect to time
22 Jan 09
Originally posted by joe shmoExactly, that's why I thought yours was wrong. Assuming y = -1.6 m (you would need to specify that value) and Y,initial = 0 meters or y = 0 meters and Y,initial = 1.6 meters.
how do you know that my hieght of 1.6m wasnt my "yo"=0, thus leaving the final position of y=-1.6m?
My solution is an alternate method. Finding the peak height is typically required in these multi-step problems. I split the problem into 3 sections: reaching the peak, falling back down to the initial height, and then falling 1.6 meters extra. Works all the same.
And greeting fellow handler. When did you start?
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22 Jan 09
Originally posted by Ramnedgood now that we've cleared that up...lol
Exactly, that's why I thought yours was wrong. Assuming y = -1.6 m (you would need to specify that value) and Y,initial = 0 meters or y = 0 meters and Y,initial = 1.6 meters.
My solution is an alternate method. Finding the peak height is typically required in these multi-step problems. I split the problem into 3 sections: reaching the peak, falling back d ...[text shortened]... lling 1.6 meters extra. Works all the same.
And greeting fellow handler. When did you start?
That had me up nights.......It is a tough thing to deal with, when someone tells you your incorrect, when every inch of your body tells you the opposite!
by package handler, do you mean driver? I'm just a lowly belt worker..🙂 I started on the pre-load, where you get get up at 3 am and work until 9 am. After work I would head off to school, but doing that wore me out, so now im on the re-load..or the local sort...I don't know how your plant is, but at ours, the job requires a great deal of physical stamina...in addition 20 hrs plus a week difficult with full time school...I see your a physics major, how are you handeling work and school?
22 Jan 09
Yep, I work loading the cars. Used to be sore every night, starting to wear off, which is a good sign. Desperately waiting to get promoted to the driver spot....physics + UPS workout is tiring!! I still have a year or two before I get that chance I think. Most people go "pfft" at UPS drivers, as an easy job but it takes a lot of work to get there.
