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Old 09-26-06, 11:00 AM   #1
phantomcow2
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Beast of a physics problem

I have to say, Physics has been going well for me. But he assigned a problem in preparation for a test. Well needless to say, I am not sure where to begin! I was hoping somebody could at least point me in the right direction.

A baseball is hit for a home run, it travels 138m. It just clears a 6.5m wall when it lands, where it is caught by a lucky fan. So the ball essentially starts off at point (0,0), and ends at (138,6.5).
It is hit at a 40 degree angle. Calculate initial velocity and total flight time.

I know it must be possible. Because only at one velocity at 40degrees will the ball actually go 138m. Like if it was hit at 1m/s, obviously it would not go the full 138.

I've got a few ideas brewing upstairs, but ugh.
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Old 09-26-06, 11:05 AM   #2
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Would it have traveled farther than 138m if the ball had returned to the level where it was hit (somewhat less than 6.5 meters off the ground I'd presume)?
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Old 09-26-06, 11:11 AM   #3
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Study Danno's solution to your last physics problem. Remember, the whole point of these problems is that you can separate out motion in the two orthogonal directions. Do that.

You have two unknowns. In this case, conveniently, they're explicitly given to you. They're flight time and initial velocity. You have two equations, one for each dimension. Voila, two equations and two variables. Solve.
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Old 09-26-06, 11:24 AM   #4
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Hmm...

I could be wrong, but here's what I'd do.

Break down the vector into vector components. You should be able to find the time it takes for the ball to reach its peak height. Then you should be able to find the time for the ball to reach 6.5m from its peak height. From there, you should have total time. Then you can find the Xi vector component. Then, use some trig magic and find Vo.

Heh, I'd see if I can work it for you, but I don't have my old notes w/ me
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Old 09-26-06, 11:40 AM   #5
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The problem as given can not be solved. You need at least the barametric preasure and the wind speed and direction.

Unless of course they play baseball in a vacuum.

I just love dorky teachers who try to make 'realistic' word problems and then forget about the real world problems involved.

BTW size and weight of the baseball also matter, and balls are NOT identical at all levels of play.
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Old 09-26-06, 11:45 AM   #6
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Originally Posted by Keith99
The problem as given can not be solved. You need at least the barametric preasure and the wind speed and direction.
Are you one of those people that when asked "What book would you bring if you were trapped on a desert island?" would respond "Forget the book, I'll need water, food rations, a compass and a raft"

???
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Old 09-26-06, 11:45 AM   #7
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Originally Posted by USAZorro
Would it have traveled farther than 138m if the ball had returned to the level where it was hit (somewhat less than 6.5 meters off the ground I'd presume)?
Oops forgot that part. Even if this is in a vacuum you need to know the height of the original contact.
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Old 09-26-06, 11:47 AM   #8
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Originally Posted by Keith99
Oops forgot that part. Even if this is in a vacuum you need to know the height of the original contact.
I think you could safely assume 1m.
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Old 09-26-06, 11:47 AM   #9
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Separate this into two steps... Then the same process on each as before, separate vertical from horiontal...

Last edited by DannoXYZ; 09-26-06 at 03:28 PM.
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Old 09-26-06, 11:47 AM   #10
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Originally Posted by Crono
Are you one of those people that when asked "What book would you bring if you were trapped on a desert island?" would respond "Forget the book, I'll need water, food rations, a compass and a raft"

???
Nope, I'm the one who picks a book on how to get off a desert island.
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Old 09-26-06, 11:57 AM   #11
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Originally Posted by Keith99
The problem as given can not be solved. You need at least the barametric preasure and the wind speed and direction.

Unless of course they play baseball in a vacuum.

I just love dorky teachers who try to make 'realistic' word problems and then forget about the real world problems involved.

BTW size and weight of the baseball also matter, and balls are NOT identical at all levels of play.
A-lass do not forget to factor in humidity (drag) and the ball stiching thread diameter.....(more drag)
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Old 09-26-06, 12:15 PM   #12
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Originally Posted by tcar5
A-lass do not forget to factor in humidity (drag) and the ball stiching thread diameter.....(more drag)
Are you implying that a ball with more pronounced stitching will have more drag and therefore not fly as far? I wouldn't. Because a back-spinning baseball would have a bit of lift due to the Bernoulli Principle. http://en.wikipedia.org/wiki/Bernoulli%27s_principle
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Old 09-26-06, 12:33 PM   #13
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LOL... I'd be a wise-arse and bring all that stuff up... Is there wind? What's the density altitude? What's the barometric pressure? Is the ball spinning? Lift causes drag too, don't forget. Factor that in.

It's like the old joke about the student who was asked how to determine the height of a building. His answer: "Ask the building manager" or something like that.
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Old 09-26-06, 01:05 PM   #14
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Back in my American Legion days when I was regularly serving up non-sliding sliders, we would describe the speed of this baseball as "serious neck." As in how fast your neck would turn watching that ball get out.
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Old 09-26-06, 03:31 PM   #15
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Quote:
Originally Posted by phantomcow2
I have to say, Physics has been going well for me. But he assigned a problem in preparation for a test. Well needless to say, I am not sure where to begin! I was hoping somebody could at least point me in the right direction.

A baseball is hit for a home run, it travels 138m. It just clears a 6.5m wall when it lands, where it is caught by a lucky fan. So the ball essentially starts off at point (0,0), and ends at (138,6.5).
It is hit at a 40 degree angle. Calculate initial velocity and total flight time.

I know it must be possible. Because only at one velocity at 40degrees will the ball actually go 138m. Like if it was hit at 1m/s, obviously it would not go the full 138.

I've got a few ideas brewing upstairs, but ugh.
Sorry man i can't show you my steps but here it is: The pitcher threw the ball 87 mph, stayed in the air about 6.73 seconds. Might be a little off but i took a try.
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Old 09-26-06, 03:58 PM   #16
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Quote:
Originally Posted by vtjim
It's like the old joke about the student who was asked how to determine the height of a building. His answer: "Ask the building manager" or something like that.
Supposedly, it was Niels Bohr:

A student refused to parrot back what he had been taught in class.
When the student protested, I was asked to act as arbiter between
the student and his professor.

I went to my colleague's office and read the examination
question: 'Show how it is possible to determine the height of a
tall building with the aid of a barometer.'

The student had answered: 'Take the barometer to the top of
the building, attach a long rope to it, lower the barometer to
the street and then bring it up, measuring the length of the
rope. The length of the rope is the height of the building.'

A high grade is supposed to certify competence in physics, but
the answer did not confirm this. I suggested that the student
have another try at answering the question. I gave the student
six minutes, with the warning that his answer should show some
knowledge of physics. In the next minute he dashed off his
answer, which read: 'Take the barometer to the top of the
building and lean over the edge of the roof. Drop the barometer,
timing its fall with a stopwatch. Then, using the formula
S = 1/2at^2, calculate the height of the building.'

At this point, I asked my colleague if he would give up. He
conceded, and I gave the student almost full credit.

In leaving my colleague's office, I recalled that the student
had said he had other answers to the problem, so I asked him what
they were.

'Oh, yes. There are many ways of getting the height of a tall
building with the aid of a barometer. For example, you could take
the barometer out on a sunny day and measure the height of the
barometer, the length of its shadow, and the length of the shadow
of the building, and by the use of a simple proportion, determine
the height of the building.'

Fine, I said. And the others?

'Yes. Take the barometer and begin to walk up the stairs. As
you climb the stairs, you mark off the length of the barometer
along the wall. You then count the number of marks, and this will
give you the height of the building in barometer units. A very
direct method.'

'Finally, there are many other ways of solving the problem.
Proably not the best is to take the barometer to the basement and
knock on the superintendent's door. When the superintendent
answers, you speak to him as follows: "Mr. Superintendent, here
I have a fine barometer. If you will tell me the height of this
building, I will give you this barometer".'

Other solutions off the net:

Walk away from the building with the barometer at arm's length.
Once the apparent height of the barometer is the same as the
building's, measure the distance from the building and the height
of the baraomter and use a little trig.

Tie the barometer to the end of a long string such that the
end just touches the ground when you hold it from the roof.
Raise it (say) one foot. Swing the barometer-pendulum and
time its period, then calculate the length of the pendulum
from the pendulum equation T = 1/(gL)**0.5 -- don't forget
to add back that foot.

Walk back a measured distance from the building. using any convenient means, throw the barometer at the top of the building. (Use trial and error until you get the aim right) Measure the angle from the ground and the initial velocity, account for wind and air resistance, use several formulae, and be prepared to account for why you just smashed the sh*t out of the professor's new barometer.

Give the barometer to a cab driver in exchange for him taking you over to City Hall so you can look up the height of the building in their records.

Use the barometer to weigh down a piece of string and lower the string down the side of the building. Then you could: a) pull the string up and measure it, or b) start it oscillating like a pendulum, measure the length of a period, and calculate the length of the string from that.
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Old 09-26-06, 04:14 PM   #17
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Assuming the intended answer is measuring the difference displayed on the barometer when used at teh top and bottom of the building and calculating the elevation difference the non-traditional methods are probably going to give a more accurate answer. (Well the building super method will depend on how honest and knowledgeable he is).
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Old 09-26-06, 05:10 PM   #18
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well first of all, we are told to neglect pressure, air resistance, all that stuff.
I've understood you need to break it down into vectors. Here is what I thought to do:
So 6.5=0 + Vot + -4.905t^2

This is for our vertical vector. So gravity has an effect. 6.5= final distance. 0=initial distance.

time is unknown, as is initial velocity. Another equation I know is
V1=V0 + at

But in this one, I only know one variable! I must be missing something here.
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Old 09-26-06, 05:17 PM   #19
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What the heck is Vot, V1, V0, at? It would help us help you if you clearly define your variables. (And don't use terms like 4.905. What the heck is 4.905? Use 1/2 g and plug in the value of g later.) I managed to wade through all that, but it took a lot longer than it should have taken.

Yes, you are missing something. Hint: In simple high school problems, you generally are given exactly the information you need. You are given no less information, but also no more information. (If only things were this way in real life!) What information have you not used in your equations? Why does that information matter? How can you use that information?
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Old 09-26-06, 05:22 PM   #20
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Sorry,
V0 is initial velocity, t is time. V0t is initial velocity * time. a is the variable for acceleration, at is acceleration*time. -4.905 is 1/2*-9.81

That is all the information given to us. Everything he has given us has been used in my equations. He has been known to give problems that don't work, just to see if we catch it. But, I am positive this one is possible.
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Old 09-26-06, 05:28 PM   #21
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Mr. Ghost Bovine, let's break down this problem, shall we?

You have a projectile moving in two axes: x (horizontal) and y (vertical). As you have already correctly identified, it starts at (x,y) = (0,0) and ends up at (x,y) = (138,6.5). Very good. You've also started writing some rudimentary equations for x(t) and y(t). In particular, since in general z(t) = z(0) + Vz*t + 1/2(az)(t^2) where Vz = initial velocity in z axis and az = acceleration in the z axis,

x(t) = x(0) + Vx*t + 1/2(ax)(t^2)
y(t) = y(0) + Vy*t + 1/2(ay)(t^2)

Furthermore, you can hopefully see that ax = 0 and ay = -g. And at time t of interest, you know the values of x(t) and y(t). You also know x(0) and y(0). That's how you can get

138 = Vx*t
6.5 = Vy*t - 1/2 g*t^2

[EDIT] Does all this make sense up to this point? If not, what doesn't make sense to you? [/EDIT]

Now... how do you relate Vx and Vy? And what else are you supposed to solve for? Oh yeah... Vtot, the initial total velocity. Can you relate Vx and Vy to Vtot? what information is given to you to relate those unknowns?
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Old 09-26-06, 05:35 PM   #22
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Ok joking about the poorly phrased problem aside. Here is my hint. Break it down into what is happening horrizontally and vertically. Remember there is only 1 ball so it is in the (magic no resistance) air for the same amount of time. Write each equation so it just describes what is happening first, then try to link them to get an answer second.

As JS said define all terms clearly and explicitly first.
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Old 09-26-06, 05:35 PM   #23
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hmm...
I keep thinking about systems of equations. I am trying to find that perfect velocity where hte ball would land at (138,6.5) when launched at a 40degree angle. Would it be possible to setup a system of equations, one equation for each vector?

6.5=0 + Vot + -4.905t^2
138=0+Vot + 0t^2

Multiply bottom equation by (-1) to invert the sign of Vo (initial velocity). Cancel that out, add the two equations together, solve for T. Plug T back into the equation?

That still seems wrong
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Old 09-26-06, 05:35 PM   #24
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Seriously, make sure you understand the entire thought process. Don't rely on rote or formula. Tell us what doesn't make perfect sense to you, and we can try explaining it a hundred different equivalent ways in hopes that one of the explanations (all of which will be mathematically equivalent) will help yo "get it". But for you to "get it", you have to let go of the crutch of relying on fitting problems into premade molds and instead try to see the big picture in the problem and understand each component of the big picture.
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Old 09-26-06, 05:38 PM   #25
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hmm...
I keep thinking about systems of equations. I am trying to find that perfect velocity where hte ball would land at (138,6.5) when launched at a 40degree angle. Would it be possible to setup a system of equations, one equation for each vector?

6.5=0 + Vot + -4.905t^2
138=0+Vot + 0t^2

Multiply bottom equation by (-1) to invert the sign of Vo (initial velocity). Cancel that out, add the two equations together, solve for T. Plug T back into the equation?

That still seems wrong
We cross posted. The reason this is wrong is because the initial velocity is not the same as the initial velocity along each vector. See my post above yours if necessary. Go back to your other thread if necessary.

By the way, there will be two mathematically correct answers. (I know this without having tried to actually solve your problem numerically.) Why is this? How do you know which one is correct?
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