If anyone's awake projectile motion

AI Thread Summary
The discussion revolves around solving a projectile motion problem involving a football thrown at an angle of 20 m/s, landing 25 meters away. The correct formula for the range of a projectile, R = (u^2 sin(2θ))/g, is highlighted, leading to two possible angles for the same range: 18.9 degrees and 71.1 degrees. Participants explain that the two angles result from the properties of the sine function, where sin(2θ) intersects a horizontal line at two points within the interval. The conversation emphasizes the relationship between angle and range in projectile motion, particularly noting that the maximum range occurs at 45 degrees. The discussion concludes with a light-hearted acknowledgment of the complexities of physics, especially for those not majoring in the subject.
Chib
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All right, before I look like a big idiot, I want to preface this with a statement that I have been through Cal III, Abstract Algebra, I'm a senior in college, and I got a 620 on the math portion of the SAT.

And I CANNOT get this basic problem in projectile motion figured out. Physics has humbled me. Anyone that understands physics gets my highest regard after just getting to the third chapter in this algebra-based physics book.

Here's my problem:

We've got a football being thrown at an undetermined angle, theta, at 20m/s. It lands (at the same height, for the sake of ease) 25 meters away.

I know that we know Xnaught, X, Vnaught, asuby, ynaught and y.

So I start trying to set the problem up and come up with:

x=-2VoxVoy/ay

I know that the Vnaughtx (which I can't type, since it looks dirty o:) ) is basically the velocity (20 m/s) times the cosine of the angle of release. Vnaughty (:blushing: ) is the same, but with sine, right?

Okay, so I solve down algebraicially to (xay)/(-2v^2) = cos(theta) sin(theta)

And I've got nowhere to go after that. I got nothin'. I checked out the trig identities, thinking that might be the equivalent of Tangent, but it's not.

Am I using the wrong equation?

Thanks in advance for anyone who can help me not fail this class.
 
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The correct equation for the range of a projectile,
R = \frac{u^2sin(2\theta)}{g}

If you look closely you will find that you get the same range for \theta and 90-\theta.
Can you do the problem now ?
If you want to know how the equation is derived, I will be happy to give you a walkthrough. :D
 
Thank goodness you were on. I looked through when I saw that and didn't see that equation anywhere in the book. I think since this is an intro physics book based on algebra, we had specific equations for specific instances.

Okay, I took it down to arcsin(Rg/v^2)/2 = \theta and got 18.9 degrees. Checked with the back of the book, and it looks like it's good. However, it also gives 71.1, which is 90-\theta, like you said. I'm not following why that's the case. Is it more of those crazy circle properties?

Again, thank you SO much for helping me out with that.
 
The max range for any given initial velocity occurs at \Theta =45 deg. Therefore there will always be 2 angles which result in the same range, one less then 45 the other greater then 45.
 
I guess that makes sense; I just can't picture it in my head.

Good thing I'm just a psych major, and most of the physics I'll do in my life will involve an open book, a pencil and a paper! ;)
 
Also, consider the sine function. Sketch a sine graph (\sin(2\theta) vs. \theta), now draw a horizontal line above the x - axis. Now how many times does this line intersect the curve in the period? The interestions are where your solutions lie. However, you calculator only shows the values at 0+ \theta, not at 90 - \theta or any subsequent values.
 
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Hoot, little correction there, maybe just a typo .
You should plot sin2\theta Vs \theta[/itex] and then draw the horizontal line to intersect at two points while \theta lies in the interval (0 , \frac{\pi}{2}) .<br /> <br /> Oh, and chib , you&#039;re most welcome .
 
arunbg said:
Hoot, little correction there, maybe just a typo .
You should plot sin2\theta Vs \theta[/itex] and then draw the horizontal line to intersect at two points while \theta lies in the interval (0 , \frac{\pi}{2}) .<br />
<br /> Thank&#039;s arun, it was a typo (otherwise we would only have one value for theta in the interval). I have duely correct it.
 
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