If anyone's awake projectile motion

In summary, The conversation discusses a problem in projectile motion involving a football being thrown at an angle of 20m/s and landing 25 meters away. The correct equation for the range of a projectile is given, and the steps to solve the problem are outlined. The conversation also mentions the use of trigonometric functions and how there can be two angles that result in the same range. The conversation concludes with a discussion on how to plot the sine function to find the solutions.
  • #1
Chib
3
0
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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  • #2
The correct equation for the range of a projectile,
[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]

If you look closely you will find that you get the same range for [itex]\theta[/itex] and [itex]90-\theta[/itex].
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
 
  • #3
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 = [itex]\theta[/itex] 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-[itex]\theta[/itex], 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.
 
  • #4
The max range for any given initial velocity occurs at [itex] \Theta [/itex] =45 deg. Therefore there will always be 2 angles which result in the same range, one less then 45 the other greater then 45.
 
  • #5
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! ;)
 
  • #6
Also, consider the sine function. Sketch a sine graph ([itex]\sin(2\theta)[/itex] vs. [itex]\theta[/itex]), 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 [itex]0+ \theta[/itex], not at [itex]90 - \theta[/itex] or any subsequent values.
 
Last edited:
  • #7
Hoot, little correction there, maybe just a typo .
You should plot [itex]sin2\theta[/itex] Vs [tex]\theta[/itex] and then draw the horizontal line to intersect at two points while [itex]\theta[/itex] lies in the interval [itex](0 , \frac{\pi}{2})[/itex] .

Oh, and chib , you're most welcome .
 
  • #8
arunbg said:
Hoot, little correction there, maybe just a typo .
You should plot [itex]sin2\theta[/itex] Vs [tex]\theta[/itex] and then draw the horizontal line to intersect at two points while [itex]\theta[/itex] lies in the interval [itex](0 , \frac{\pi}{2})[/itex] .
Thank's arun, it was a typo (otherwise we would only have one value for theta in the interval). I have duely correct it.
 

Related to If anyone's awake projectile motion

1. What is projectile motion?

Projectile motion is the motion of an object through the air or space under the influence of only gravity. It is a type of motion that follows a curved path known as a parabola.

2. How is projectile motion different from regular motion?

Unlike regular motion, which can have both horizontal and vertical components, projectile motion only has a vertical component due to the force of gravity. This means that the object will always follow a curved path.

3. What factors affect projectile motion?

The factors that affect projectile motion include the initial velocity, the angle at which the object is launched, and the force of gravity. Air resistance can also have an impact on the motion of the object.

4. How can we calculate the trajectory of a projectile?

To calculate the trajectory of a projectile, we can use the equations of motion, which take into account the initial velocity, angle of launch, and acceleration due to gravity. We can also use computer simulations to model and predict the path of a projectile.

5. What are some real-life examples of projectile motion?

Some real-life examples of projectile motion include throwing a ball, shooting a basketball or arrow, and the motion of objects in space such as planets and comets. Projectile motion is also used in sports like javelin throw and long jump.

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