Foo - 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.

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)?

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.

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 :)

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.

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"

???

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.

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.

Separate this into two steps... Then the same process on each as before, separate vertical from horiontal...

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.

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)

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

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. :D

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.

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.

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. :D

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.

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).

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.

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?

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.

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

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

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?

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.

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

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.

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?

So you are asking how do I relate the velocities in the individual vectors? I do this by drawing a right triangle. Your two sides are your X and Y vectors. Your hypotenuse is the actual speed, that is what I will need. You simply use the pythagorean theorem to determine the vector triangles hypotenuse, this is the actual speed.

So you are asking how do I relate the velocities in the individual vectors? I do this by drawing a right triangle. Your two sides are your X and Y vectors. Your hypotenuse is the actual speed, that is what I will need. You simply use the pythagorean theorem to determine the vector triangles hypotenuse, this is the actual speed.
Think some more. Yes, I am asking you to relate the velocities in the individual vectors. Yes, the hypotenuse is the actual initial velocity. You're missing something. What other information were you given? How do you use it?

PS: I'll be surfing other subforums. PM me when you make any more progress or feel stuck (don't have to write anything, just make sure I get a PM from you) and I'll come checking on this thread.

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?

There will be two values for T becuase the quadratic equation is necessary, which always gives 2 answers. One will often be negative. Well you can't have negative time, so I assume not use that.

The only time I have used both is when I have a problem like
"A cat is fired from a cannon at 55 degrees @ 151m/s. Calculate the time when the cat is at 15m of height."
Then you will have two positive times. Because one is for the "climb". Like as the cat reaches the fulcrum point in his parabolic journey. Then when he is at 15m height on his descent.

Very good. Now get back to the problem at hand.

oh! You must be talking about the angle at which the ball was projected! That certainly relates the two vectors.

The way they are really related would be using the tangent function, opposite over adjacent.
Vy/vx = Tan(40

Is this what you were talking about?

Here's a hint... how much time would it take the ball to fall 6.5m??? That's another given in the problem...

and again... no need to use quadratics... drawing a picture and filling in all the given variables really help you set up the equations. Then solve them. :)

oh! You must be talking about the angle at which the ball was projected! That certainly relates the two vectors.

The way they are really related would be using the tangent function, opposite over adjacent.
Vy/vx = Tan(40

Is this what you were talking about?
Yes!

By the way, DannoXYZ's suggestion is an excellent one. If you cannot visualize a problem instantly, drawing a good diagram is usually an excellent way to begin.

I seem to remember something in Physics I about using the time-based function for vertical distance divided by the time-based function for horizontal distance to get the position of the object in flight at any point in time, but I thought it was more complex than tan(time)=height(time)/distance(time).

Does that make sense to anyone?

Mr. Ghost Bovine, now that you understand this problem, let's see how well you understand it. Why don't you explain to TCNJCyclist what he's doing wrong?

I wish I did understand the problem :(. Well, I've gotten far I think.
DannoXYZ. Are you asking how long it would take to reach the ground if I simply dropped a ball from a height of 6.5m? Or how long would it take to go that extra 6.5m after it has gone through its trajectory?

jschen, you are an admirable soul. :)

I wish I did understand the problem :(. Well, I've gotten far I think.
Keep at it. You are doing a lot better with it now than you were in post #1.

Keep at it. You are doing a lot better with it now than you were in post #1.
Thanks for the motivation :). The "missing link" was solved, that is the link between our two vectors (tangent function). Now, the mystery is what on earth am I going to do with this slab of information?

Now, the mystery is what on earth am I going to do with this slab of information?

:lol: Well, based on our last conversation, at this point, you'd break out the calculator.

But let me help some more before we get all computational. Rather than have a plethora of variables, let's express things in terms of what we actually care about: initial velocity (Vtot).

Vy = Vtot sin 40 deg
Vx = Vtot cos 40 deg

Does that help any?

I just pulled out my old Physics I book and found the trajectory equation, and yeah, I was wrong.

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

Think mortars and trajectories. The ball travels alot father in it's arc than the distance out from contact.
Don't forget gravity taking away and then adding to the equations

JS,

In a previous thread you were wondering about what you should do professionally. After reading this thread I think teaching does have a place in your plans. That is NOT to say it should be your primary profession. But I think you should seriously look for some way to include some teaching. Teaching one class at a J.C. comes to mind. Or perhaps doing some teaching after you have made the big bucks.

PCOW. Some Jr. High advice, but advice that was echoed in my college classes also. For any work problem when you are done plug the answer back in and see if it makes sense. In the case of this one you might be surprised at how many people include both answers and get dinged a few points for it. It is also a good way to catch errors like dropping a decimal place.

:lol: Well, based on our last conversation, at this point, you'd break out the calculator.

But let me help some more before we get all computational. Rather than have a plethora of variables, let's express things in terms of what we actually care about: initial velocity (Vtot).

Vy = Vtot sin 40 deg
Vx = Vtot cos 40 deg

Does that help any?


I understand where you get this from, but I am not sure what to do with this information. This has been root of the problem from square 1 I think. :(

phantomcow2, you have two unknowns that you ultimately want to solve for: t and Vtot. So far, you've got equations with the following variables: t, Vx, and Vy. By substituting Vtot-based terms for Vy and Vx, you now have two equations and two variables. Now it's just a big math problem.

I had a feeling I was substituting...
I wrote all of this down. Tomorrow, I will solve this math problem when my mind is fresh after a quality 8 hours of sleep.
Thanks very much for the help

By the way, you could have chosen to express Vx in terms of Vy (or vice versa) using the tangent function as you were doing earlier. You'd still end up with two variables and two equations. I just prefer my way because you directly solve for what you actually care about, rather than having to obtain an intermediate answer before going back and actually solving for what you care about. The two methods are mathematically equivalent. But with practice, you can learn to focus on what you care about and start expressing things in terms that allow you to get the answer you want directly. It's more conveninent, and at least in my brain, it makes for more straightforward thinking.

Keith99, you have a scary good memory.

I wish I did understand the problem :(. Well, I've gotten far I think.
DannoXYZ. Are you asking how long it would take to reach the ground if I simply dropped a ball from a height of 6.5m? Or how long would it take to go that extra 6.5m after it has gone through its trajectory?There's no "extra" 6.5m that the ball moves. It's actually a clue and is an integral part of the solution. This problem is actually the reverse of the last one. Previously, you were given velocity and had to figure out time and distance. In this problem, you were given distance, just work backwards and find out time, then the velocity required to cover that required distance in the computed amount of time.

Draw a picture, it'll make A LOT more sense. :)