Two Dimensional Motion Projectile

1. Mar 24, 2004

SSVicious

Hi. I haven't done physics for over a year and a half now, but my friend is taking the class. He asked for my help, but I really didn't know how to help him. Here are the questions...

1) Show how Delta y = (1/2)* (tan theta) * delta x
and why Delta y= (1/2) * delta y prime

2) Show why delta x is maximized for a given velocity when theta is equal to 45 degrees.

This is two dimensional motion. Here is the diagram:

Delta y Prime
| /\
| / \
| / \
| / \
| / \
| / \
|/ \
|-------------------------------
Delta X

The projectile motion is a line going from the beginning of the triangle, to the center of it, to the end of it. Delta Y is in the middle of Delta y prime and the middle of the triangle.

2. Mar 24, 2004

Integral

Staff Emeritus
more information concerning the level of the class and available mathematical tools would help bring meaningfull responses.

Are there any other assumed equations not given in the problem? For example some expression for y?

3. Mar 24, 2004

SSVicious

It's an introductory physics class that only covers 1 dimensional motion, 2 dimensional motion, and circular motion. They are doing 2 dimensional motion. It is a simple physics course, but alas I do not remember how to do the stuff. Here is a drawing attached to this post.

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4. Mar 24, 2004

SSVicious

There, I posted an image of it.

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5. Mar 24, 2004

ShawnD

It took me half a page of paper to show this lol. My writing on paper is correct but I may make a few errors typing it out.

First make a big formula for distance then take the derivative with respect to theta.

delta x:

$$x = V_xt$$

$$x = Vcos(\theta)t$$

now find what t is
vertical velocity, when the object is at vertical maximum:

$$V_{yf} = V_{yi} + \frac{at}{2}$$

$$0 = Vsin(\theta) + \frac{at}{2}$$

$$t = \frac{2Vsin(\theta)}{a}$$

now fill that into the previous equation

$$x = Vcos(\theta)(\frac{2Vsin(\theta)}{a})$$

$$x = \frac{2V^2sin(\thetea)cos(\theta)}{a}$$

$$x = [\frac{2V^2}{a}] [sin(\theta)cos(\theta)]$$

$$\frac{dx}{d \theta} = [\frac{2V^2}{a}] [cos(\theta)cos(\theta) + (-sin(\theta))sin(\theta)] = 0$$

As long as the object was thrown and the earth has gravity, 2v^2/a will never be 0 so we can just ignore it completely.

$$0 = cos^2(\theta) - sin^2(\theta)$$

$$0 = cos^2(\theta) - (1 - cos^2(\theta))$$

$$0 = cos^2(\theta) - 1 + cos^2(\theta)$$

$$2cos^2(\theta) = 1$$

$$cos^2(\theta) = \frac{1}{2}$$

$$cos(\theta) = \frac{1}{\sqrt{2}}$$

$$\theta = 45$$ degrees

Last edited: Mar 24, 2004
6. Mar 24, 2004

SSVicious

Wow, that was a lot. But extremely helpful. Thanks for help on question 2.

7. Mar 24, 2004

ShawnD

Can you explain a little more in depth what the difference between Y and Y prime is? Y prime is the top and Y is half of that or something?

First expand the formula to look for anything familiar

$$Y = \frac{sin(\theta)X}{2cos(\theta)}$$

One thing I notice is X/cos(theta). That is the same as the object's trajectory which I will call R.
Another thing is that bringing the sine to the other side will create another R

$$\frac{Y}{sin(\theta)} = \frac{R}{2}$$

$$R = \frac{R}{2}$$

If Y is actually supposed to have half of Y prime (which I used in the equation), that would make the equation work.

Last edited: Mar 24, 2004
8. Mar 24, 2004

SSVicious

all i know is that y prime is the derivative of y.

9. Mar 24, 2004

ShawnD

Ok then....

When you say delta Y do you mean Y or do you mean dY like it's a differential?

10. Mar 24, 2004

SSVicious

delta y as in the change of y, not the derivative.