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Science > Physics |
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"" |
| Date: |
28 Feb 2007 03:31:34 PM |
| Object: |
physics help |
for the non-physics minded person?
ive done the problem, i just dont know what certain things stand for,
and i dont quite understaND WHY THE ANSWER WAS WRITTEN THE WAY IT
WAS. im sorry i DID NOT mean to yell. caps lock got hit.
problem and my method appears here.
http://www.topix.net/forum/science/physics/TMCNC9KHL8NL2V8HR
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| User: "Greg Neill" |
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| Title: Re: physics help |
28 Feb 2007 04:11:17 PM |
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<fest_jam@yahoo.com> wrote in message
news:1172698294.016746.91300@8g2000cwh.googlegroups.com...
for the non-physics minded person?
ive done the problem, i just dont know what certain things stand for,
and i dont quite understaND WHY THE ANSWER WAS WRITTEN THE WAY IT
WAS. im sorry i DID NOT mean to yell. caps lock got hit.
problem and my method appears here.
http://www.topix.net/forum/science/physics/TMCNC9KHL8NL2V8HR
From your web page the problem is as follows:
A projectile is fired straight upward from ground level
with an initial velocity of 128ft/sec.
A) At what instant will it be back at the ground?
B) When will the height be less than 128 feet?
It's a typical projectile problem. Known items:
1. Acceleration due to gravity is -32 ft/second^2
2. Initial velocity: 128 ft/second
3. Your choice of coordinate system has the vertical direction
positive, which is why the acceleration is negative (it
points downwards). The origin is at ground level.
You've used the formulae:
v(t) = v_0 + a*t (note: subscripts indicated by underscore)
x(t) = x_0 + v_0*t + (1/2)*a*t^2
Where x_0 is the initial height (Ground level, so x_0 = 0)
v_0 is the initial velocity, 128 ft/sec
When you solved part A you chose to solve for the time when
the projectile reached apogee (the top of its trajectory)
and then used symmetry, doubling the value to find the total
time for the projectile to rise and fall back to the ground.
That works. You could also have used the second formula to
determine the times when the height of the projectile is 0 ft.
To solve part B you need to realize that the projectile will
be at a height of less than 128 feet during two portions of
its trajectory: on the way up (from the groud to 128 ft) and
again on the way down (from 128 ft back to the ground). So
the corresponding times will be time ranges, usually expressed
as 0 <= t < t_1 and t_2 < t <= t_3, where you need to determine
the values of t_1, t_2, and t_3 (hint: you found t3 previously
when you calculated the total time of the projectile flight).
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| User: "Sam Wormley" |
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| Title: Re: physics help |
28 Feb 2007 03:56:31 PM |
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wrote:
for the non-physics minded person?
ive done the problem, i just dont know what certain things stand for,
and i dont quite understaND WHY THE ANSWER WAS WRITTEN THE WAY IT
WAS. im sorry i DID NOT mean to yell. caps lock got hit.
Trajectory
http://scienceworld.wolfram.com/physics/Trajectory.html
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| User: "PD" |
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| Title: Re: physics help |
01 Mar 2007 09:58:45 AM |
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On Feb 28, 3:31 pm, wrote:
for the non-physics minded person?
ive done the problem, i just dont know what certain things stand for,
and i dont quite understaND WHY THE ANSWER WAS WRITTEN THE WAY IT
WAS. im sorry i DID NOT mean to yell. caps lock got hit.
problem and my method appears here.
http://www.topix.net/forum/science/physics/TMCNC9KHL8NL2V8HR
Couple of comments:
For part (a) the way that it was written is fine, but is not the only
way to do it.
If it is fired upward at 128 ft/s (v_0 = +128 ft/s), then when it
comes back down to earth it should be going 128 ft/s downward (v_f =
-128 ft/s). Notice how the signs keep track of the direction: we're
saying up is + and down is -. It doesn't matter as long as you're
*consistent*.
Then you can use that same formula:
v_f = v_0 + a*t
-128 ft/s = +128 ft/s + (-32 ft/s^2) * t
(Do you see why there's a - in front of 32 ft/s^2? What direction is
that acceleration?)
Now when you solve for t, you'll find 8 seconds directly.
For part b, think about what's being asked. It's asking you for *what
time* is the vertical *position* less than 128 ft.
First thing to draw is a picture. It starts off at 0 ft, goes up,
blows past 128 ft, reaches a peak higher than 128 ft, comes back down,
blows past the 128 ft mark again, and then hits the ground at 0 ft
again. So on your trajectory, mark 5 points: where it starts (point
1), blowing by 128 ft on the way up (point 2), the point at the top
(point 3), blowing by 128 ft on the way down (point 4) and hitting the
ground again (point 5). Now it's plain what it's asking for. It's
asking for the all the times between points 1 and 2, and all the times
between points 4 and 5. That's a shaded part of the trajectory, which
is why the answer is written as two *ranges* of t (that's what
inequalities represent: ranges). To specify the ranges, you only need
to know the endpoints of the ranges. Well, you know two endpoints:
point 1 is at 0 s, and point 5 is at 8 s (you just found that in part
a). Now you need to find the times at points 2 and 4.
So you set it up using the distance formula: x = x_0 + v_0 * t +
(1/2)a * t^2. What you are asking here is the following: If I start
out at 0 ft (that's x_0) and head upwards at +128 ft/s (that's v_0),
then at what time (that's t) will the position (that's x) be +128 ft?
Put everything in that you know and look for the thing you don't.
128 ft = 0 ft + (128 ft/s) * t + (1/2) * (-32 ft/s^2) * t^2.
OK, this looks promising. There's only one variable we don't know (t)
and that's what we wanted to know. Here's where the miracle is. Notice
this is a quadratic equation. Quadratic equations have the feature
that they generate TWO solutions. Which one do you want? Answer: you
want both! This one equation gives you BOTH answers you're looking
for. Remember, there's a time when the projectile is at 128 ft going
UP (point 2), and there is another time when the projectile is at 128
ft going DOWN (point 4). You need both, and this one equation gives
you both with one fell swoop! Isn't that slick? This is true in
general: when you have a quadratic equation and two solutions pop out,
it's always worth asking what the other solution represents -- it may
be useful.
PD
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| User: "Puppet_Sock" |
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| Title: Re: physics help |
28 Feb 2007 03:49:07 PM |
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On Feb 28, 4:31 pm, wrote:
for the non-physics minded person?
ive done the problem, i just dont know what certain things stand for,
and i dont quite understaND WHY THE ANSWER WAS WRITTEN THE WAY IT
WAS. im sorry i DID NOT mean to yell. caps lock got hit.
problem and my method appears here.
http://www.topix.net/forum/science/physics/TMCNC9KHL8NL2V8HR
If you call upwards the positive direction, then acceleration
of gravity is downwards, and so negative.
If you call upwards the "x" direction, then the equation of
up-and-down motion is
x = x0 + V0 t + 1/2 a t^2
x0 is the position at t = 0, that is the position at time zero.
In the problem it is zero.
V0 is the speed at t=0. In the problem it is +128 ft/s, positive
meaning upwards.
x is the position at time t.
All clear now?
Socks
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| User: "Matthew Lybanon" |
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| Title: Re: physics help |
28 Feb 2007 05:04:47 PM |
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in article 1172698294.016746.91300@8g2000cwh.googlegroups.com,
at wrote on 2/28/07 3:31 PM:
for the non-physics minded person?
ive done the problem, i just dont know what certain things stand for,
and i dont quite understaND WHY THE ANSWER WAS WRITTEN THE WAY IT
WAS. im sorry i DID NOT mean to yell. caps lock got hit.
And the reason you did not go back and correct your error (since you noticed
it) was . . . ?
problem and my method appears here.
http://www.topix.net/forum/science/physics/TMCNC9KHL8NL2V8HR
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