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Fostire
2010-02-21, 06:13 PM
So I've spent the last 4 hours trying to figure out a particular physics problem and no matter what i do I just can't get the correct answer.

The problem is as follows:
A cannon fires a projectile vertically upwards and the projectile reaches a maximum height H. If the angle at which this same cannon launches the projectile is changed, what is the maximum range for the projectile when launched over a horizontal surface?

The correct answer is supposed to be 2H, but I just don't understand how to get that.

HALP!

Astrella
2010-02-21, 06:18 PM
Well, don't have time to type out a complete answer now, but a way of solving it would be to try and write out the horizontal range in function of the angle, and then horizontal range. (deriving with respect to the angle, and setting the derivative to zero.) Then you can put this angle in your equation for the horizontal range to get the maximum range.

Manga Shoggoth
2010-02-22, 06:41 PM
Part of the answer is that the maximum range is achieved when the cannon is at 45 degrees to the vertical.

The range depends on two things - muzzle velocity and angle. I suspect that you need to derive the muzzle velocity from the example, then use the standard formulae to get the range.

I had a quick look (it's in my old A-level physics notes), but it's monday and I really am not in the mood for derivations. If the question isn't urgent I can have a look later in the week.

Arti3
2010-02-22, 07:14 PM
Play Portal; it will teach you everything you need to know :)

Capt Spanner
2010-02-22, 08:47 PM
From a guy with a physics degree:

For simplicity, assume no air resistance.

The cannon ball leaves the muzzle of the cannon with an intial velocity U and an elevation E. If the cannon is fired directly up in the air, the elevation is E = 90deg.

U has a vertical, and a horizontal component:

U-vert = U cos E
U-hor = Usin E

Let us consider vertical motion only. This is affected by gravitational force, which applies a constant accelleration g.

The velocity after time t after firing is therefore:

V-vert(t) = U-vert + g t [1].

To get the height of the shot at time t, integrate with respect to t:

h(t) = t U-vert + 1/2 g t^2

The maximum height, H is found when V-vert = 0. By setting this, and solving the eqn for V-vert for t, we find that maximum height is attained at time:

t(h = H) = - U-vert/g.

Substitute this in for t in the height eqn and obtain:

h-max = - U-vert^2 /2g

At 90deg, U = U-vert and h-max = H

We can also obtain a flight time, T, by solving the eqn for h, to find the value for t at which h = 0 (use the quadratic equation, with a = 1/2 g, b = U and c = 0)

T = U-vert +/- U-vert / g.

This gives solutions of:

T = 0 and T = - 2U-vert / g. [2]

Because there are no horizontal motions, the range, R, is given by:

R = T U-hor

Substituting in for T, then U-vert and U-hor we can express the range in terms of U, E and g:

R = - (2U^2 cos E sin E) / g.

You want the maximum range, which is found when E = 45deg (you can prove this by differentiating wrt E, and setting the differential to 0. You should get:
dR/dE = -(U^2 / 2g) (cos^2 E - sin^2 E).

(This is 0 for sin^2 E = cos^2 E).

Substitute in E = 45deg (so cos E = sin E = 2^-0.5 :

R-max = - U^2 / g

Compare with eqn for H:

H = - U^2 /2g

And you see that R-max = 2H. Q.E.D.

(If you don't understand any of this, I'd be happy to clarify.)

[1] This is often given as V-vert(t) = U-vert - gt, but the two forms are identical, defining positive accelerations to be in different directions: my way uses g = - 9.81 m/s/s, the other way uses g = + 9.81 m/s/s.

[2] Remember g = - 9.81 m/s/s, so we don't get negative times with this one.

Fostire
2010-02-22, 09:53 PM
The maximum height, H is found when V-vert = 0. By setting this, and solving the eqn for V-vert for t, we find that maximum height is attained at time:

t(h = H) = - U-vert/g.

Substitute this in for t in the height eqn and obtain:

h-max = - U-vert^2 /2g

At 90deg, U = U-vert and h-max = H
Ah, so that's how you get rid of that pesky t. I actually tried to do something along those lines (using the velocity=0 at maximum height) but I was trying it on the shot at an angle so it didn't really work.

(If you don't understand any of this, I'd be happy to clarify.)
The names you used for the variables are different than the ones I'm used to, but it was still pretty clear. Very helpful too.

Thank you very much :smallsmile:

Dancing_Zephyr
2010-02-22, 10:47 PM
At 45 degrees the cannonball will achieve maximum distance. Give that, you can basically draw 2 isosceles triangles to see maximum distance. But because the triangles are isosceles the base of two of them is double their height.

Manga Shoggoth
2010-02-23, 04:25 AM
From a guy with a physics degree:

A masterly summary.

Oh well, it saves me brushing the dust of my degree (and more to the point, my old A-Level revision book, which has a simillar derivation on page 8 but in much smaller print).

On the minus side, I now have nothing to do this morning while I wait for them to finish checking the power supplies to the area. Ruddy power companies... There's never a projectile around when you need one.

rakkoon
2010-02-23, 05:09 AM
:smalleek:
What?
:smalleek:
why don't you guys never ask a literature question ....

Ravens_cry
2010-02-23, 05:13 AM
:smalleek:
What?
:smalleek:
why don't you guys never ask a literature question ....
Because it wouldn't be a question so much as a discussion, because literature is art and art is about interpretation. And with interpretation there is no right answer, only a well presented one.

Manga Shoggoth
2010-02-23, 05:21 AM
:smalleek:
What?
:smalleek:
why don't you guys never ask a literature question ....

Because in maths and physics there is generally only one correct answer, and if you don't know how to get to it you are stuck.

In literature there are a range of answers and you have more of a choice. (As my old english teacher used to say, "It's OK to dislike Shakespeare, as long as you can say why you dislike Shakespeare.")

(When I was at school, my maths teacher had a degree in english, but swapped over to maths because he preferred a subject where there was one straight answer to the question...

...And, of course my english teacher had a degree in maths, and swapped to english because he preferred as subject where there wasn't.

Mind you, they were probably my favourite teachers - even the one in the subject I hated)

RebelRogue
2010-02-23, 10:23 AM
Seriously, with problems as specific as this one, I'm sure there is an equally valid "right answer" to a litterature question of similar nature.

Brother Oni
2010-02-23, 05:27 PM
The problem is that literature is less about how and more about why.

For example, in the play Macbeth, how did Macbeth become king? He murdered the old king in his sleep and framed two minions for it, nice and simple.
If you ask why he became king, then it becomes a multi-layered answer involving his wife, some witches, his greed and probably several other factors that I've forgotten.