Baseball Physics

[kid bever]

a ball is struck while traveling 50mph. It's launch angle is 30 deg. Using constant for g= 32ft/s/s calculate the distance the ball traveled. I have the answer. It's 316 ft, according to my professor. What I don't understand is this: When converting 50mph to ft/s you multiply by 1.467. Then in the equation you square that number and divide by g (32). BUT he squared 1.467 FIRST, then multiplied by 50. which gives a different answer than if you convert to ft/s first and THEN square.

 

The formula I'm using reads like this

 

R= V_i²sin(2θ)

g

 

R= (Velocity initial)² times the sine of (2*the angle of launch (or theta))

----------------------------------------------------------------------------------

32 ft/s/s

 

SO I converted 50mph to 73.35 ft/s and squared that. BUT MY PROF squared 1.467 to 2.15 THEN multiplied by 50 to get 108.

 

I hope that looks right. I know this isn't difficult math, but I just don't understand why he squared the conversion factor RATHER THAN doing the conversion and squaring the answer.

 

Thanks

 

Bruce

[mooeypoo]

I usually look at everything with the units, it is easier to understand... let's just do this step by step and see which method is right.

 

You start from 50mph( which is mi/h) and you want to get ft/s.

 

Basically, however, whatever you square, you also square the UNITS. So if you square miles/hour you will get something that's miles^2 per hour^2.. units go through the same math the values go through. For you to end up with only "Miles" at the end, you need to multiply by hours^2 and cancel those out of the denominator.

 

About unit conversion in general, here's a useful tip:

Whenever you have unit conversion, I use the fraction method. First, I will convert the miles to feet, and then the hours to minutes and then seconds. Every time I make a conversion, I multiply by the relation I want, and I take care to put the original unit on the "opposite" side of the fraction - that is, "miles" are on the numerator, I want to convert them to "feet", so I will put the relation "miles/feet" with the miles on the denominator, and hence feet on the numerator -- mathematically, it will cancel out the units.

 

So, here's the full method *you* did (first converting, then squaring):

 

[math]\frac{50 mi}{h} * \frac{5280 ft}{1 mi} = \frac{264000 ft}{h}[/math]

 

[math]\frac{264000 ft}{h} * \frac{1 h}{3600 s} = 73.34 \frac{ft}{s}[/math]

 

Which is the same as multiplying the original 50 by 1.467; but I'm just demonstrating the units. Now it's in proper units; it's the velocity.

 

[math]R = \frac{v^2(\frac{ft^2}{s^2}) \sin{60} }{g ( \frac{ft}{s^2})} = \frac{(5377.78 \frac{ft^2}{s^2}) * (\frac{\sqrt{3}}{2})}{32 (\frac{ft}{s^2})} [/math]

 

[math]R = \frac{(5380 \frac{ft^2}{s^2}) * (\frac{\sqrt{3}}{2})}{32 (\frac{ft}{s^2})}=145.54 ft[/math]

 

Your units work; the s^2 on the denominator cancel out the s^2 on the numerator; the ft in the denominator cancel only one power from the ft^2 in the numerator, and you're left with "ft", which is what R is. I don't get your answer, though.

 

 

 

This should be the right way of doing it, though. Your professor squared the "conversion factor" 1.467; units-wise, that's

 

[math]\left( \frac{5280 (\frac{ft}{mi})}{3600 (\frac{s}{h})} \right)^2 = (1.467 \frac{(\frac{ft}{mi})}{(\frac{s}{h})})^2 = 2.15 (\frac{ft^2 h^2}{mi^2 s^2}) [/math]

 

If I understand you correctly, he now took this number, multiplied by 50 (which is miles per hour) and used it as his v^2:

 

[math]v^2 = 2.15 (\frac{ft^2 h^2}{mi^2 s^2}) * 50 \frac{mi}{h}) = 107 \frac{ft^2 h}{mi s^2}[/math]

 

(as you can see, not all units cancelled out, and when we insert it back into the original equation, we get

 

[math]R = \frac{107 (\frac{ft^2 h}{mi s^2}) * \sin(60)}{32 \frac{ft}{s^2}} = \frac{9915.13 \frac{ft^2 h}{mi s^2}}{32 \frac{ft}{s^2}} = 309.85 \frac{ft*h}{mi}[/math]

 

Which make no sense in terms of units.

 

Unless I misunderstood you, it seems that squaring the conversion factor *WITHOUT* squaring the velocity and using it as-is is wrong. The initial method you used is right.

 

I seem to be getting slightly different answers, so just go over my calculations and make sure i didn't miss anything (I'm human) but the units don't lie. Conversion factor is used before you square; you square VELOCITY - conversion factor is there to help you use proper units.

 

For the record, proper units are meters and kilograms and seconds. Makes all problems so much easier...

 

~mooey

[imatfaal]

I would go along with Mooey that this is a good example of why we should stick to metric units; I would also agree with Mooey's answer. Although I would question giving to two dp when you are using 32 f/s/s - my estimate of the errors inherent mean that you cannot be sure of even the unit figure let alone decimals.

 

The prof seems to have made a clear mistake by failing to square the velocity .

[kid bever]

Mooey~

 

I follow your math perfectly. I got 145 feet as well. My professor, however ended up with 316 feet... somehow. I have no clue how.

 

Can you tell me how to illustrate the math the way you did, so I can more clearly demonstrate what he did?

 

Thanks

Bruce

 

OK, I looked more closely at the math you used to emulate my instructor, and yes that's exactly what he did. Although he rounded his number 108 rather than 107 which I think is what ends up giving me 316ft and you 309ft.

 

FYI, we normally do use SI, but since this was a baseball question he was using it to throw us "a curve" of sorts. (bad pun intended).

 

can you now illustrate flight time (t_f)using the same information? He came up with 1.56 seconds and I got 2.29 Flight time, is straight up algebra... I just multiplied all the variables, seemed pretty straightforward.

 

2times initial velocity over local gravity (32ft/s/s) times the sine of the launch angle (30 deg).

 

73.35 ft/s x two = 146.7ft/s

divided by 32 ft/s/s = 4.584s

times cosine of 30 (.5)= 2.29

 

Where in the heck did he get 1.56s ?? Am I missing something here?

[mooeypoo]

Okay, you're using

 

[math]t= \frac{2 v}{g*sin(30)}[/math]

 

All units must be unified, so if you use YOUR velocity (73.34 ft/s):

 

[math]t = \frac{2 * 73.34 (ft/s)}{32 (ft/s^2) * sin(30)}= 2.29 \frac{ft}{s} \frac{s^2}{ft}=2.29 s[/math]

 

 

Any other variation of the units like your professor got with his velocity will result in nonsensical units. Unless I'm missing something here, you professor did the initial conversion wrong and then just continued to plug in the wrong values everywhere, resulting in wrong answers.

 

Show it to him with units, it's a common way of CHECKING yourself -- it should be a good enough way to check your professor

 

 

~mooey

[kid bever]

Ok... Thanks. At least I know I'm right. It sucks as a first year, non physics major, to stare at a "simple" problem for two hours only to discover that I was right in the first ten minutes. (Hair ripping out).

 

I'm in class on Tuesday. I'll bring it up then.

 

Best

Bruce

 

 

[mooeypoo]

It sucks for the first moment of realization, and then its just awesome to know you understand how things work so well, you found an error

 

Always look on the bright side...

[kid bever]

So during the review, he realized that he had made that mistake and he was pretty flustered. THere were other people that had come to the same conclusion that I had. Thanks again for your help!