## Radar Gun Math LabColin c., Seth w., Andre D.

Our Task: We will need to find who can throw the hardest out of all of us. But, there is only one problem, the radar gun is broken.

"I think I can throw the hardest" -Seth

Early in the morning Andre, Colin, and I tested who can throw the hardest. We went to the practice football field, and all took turns throwing a baseball as hard as we could. Now, we need to find a way to calculate who actually threw the hardest by finding our miles per hour. For this we had to learn how to do the math.....

Doing the Math (Andre)

All we needed to do these calculations was our carry time, and horizontal carry (in yards). This is our data.

Andre: Andre is 5'4, weighing in at a whooping 115lb. He is currently playing a season of baseball for the the Littlestown Bolts (Junior Varsity). In the winter he also plays basketball. This is how Andre's throw went.

Andre needed to calculate how fast his throw went. This is how he did his work.

This is an explanation to how he did his calculations:

Vertical Velocity: I used the formula y= ax^2+bx+c. a in this equation represents half of the acceleration due to gravity. We chose -32.2 ft/sec as our acceleration due to gravity, so half of that is -16.1, that's how I got a. B is your vertical velocity, and since that is what we are trying to find, you have to keep it as a variable. C represents the initial hieght, so we measured where my are would be at the release point and got 5.5 feet. That is how I got the equation y= -16.1x^2+bx+5.5. Then, I substituted x and y for a point that we knew. We knew that when the ball hit the ground y would be 0. So we found the point where the ball hit the ground and that was at 1.67 seconds. So Y=0 and X=1.67 and I got the equation 0= -16.1(1.67^2)+b(1.67)+5.5. Lastly, I solved this equation and got b=23.59. SO my vertical velocity was 23.59 ft/sec.

Horizontal Velocity: I knew that to find horizontal velocity all you had to do was take the distance/time. My distance was 38 yards, but I needed it in feet. Since there are 3 feet in 1 yards, I multiplied 38 by 3 to get 114 feet. Also, my carry time in this experiment was 1.67. Knowing all of this, I got the equation 114/1.67, and after doing the math with that, I got 68.26 ft/sec.

Total Velocity: To find Total Velocity, you have to take your Vertical and Horizontal velocity, and use pythagorean theorem to find your total velocity. So I did 23.59^2+68.26^2=C^2. After going though and solving it, I got the answer 72.22ft/sec.

MPH: Once you find your total velocity, you have to convert it to MPH. To convert it you need to start by putting your total velocity over 1. So as it is in the table, 72.22ft/1 sec. Then, we are just going to make a bunch of giant 1s to cancel out the labels to get the one you want. So I put, 1 mile/5280ft, 60 sec/1 min, 60 min/ 1hr. As you can see, the seconds, feet, and minutes all cancel out. Leaving you with Miles on top, and Hours on the bottom. Then, you have to multiply all the numbers on top together, and multiply all of the numbers on the bottom together. Lastly, you take those 2 numbers (259,992mi/5280hr) and divide the top number by the bottom number, and get your MPH. After doing this, we got 49.24 MPH for Andre's Throw.

Doing the Math (Seth)

For doing these calculations for my own throw all I needed was my data on my carry time and my horizontal carry.

Seth: Seth is 5'5 weighing in at a massie 112.8 lb. Seth plays soccer in the fall for the LHS Boys Soccer team, and enjoys running track in the spring for the LHS Track and Field team. This is how Seth's throw went.

This is my work of how I did my calculations:

This is how I did my Calculations:

Vertical Velocity: First you need to find the vertical velocity that you threw. You need to set up a quadratic that that represents this. (ax^2 + bx + c) A always has to be -16.1. Your x value for the equation is your carry time, so 1.51. You need to solve for b. C is always your initital height, so 6. When I did all of this my equation looked like, -16.1(1.51^2) + b(1.51) + 6. Then I solved the equation to get a b value of 20.34. Now I need to solve for horizontal velocity.

Horizontal Velocity: Horizontal velocity is always distance thrown over time. So for me it is 99ft. over 1.51 sec. Which gives me a horizontal of 65.56.

Total Velocity: For total velocity you need to set up your vertical and horizontal velocities in a pythagrean therom way. So what I did was use this equation, 20.34^2 + 65.56^2 = C^2. (C is what we are solving for) when you solve that completley you will get a total velocity of 68.5 feet per second. Now we need to convert that data into miles per hour.

Converting to MPH: COnverting to mph, you need to make a table. First line up your total velocity with 1 sec, then line up 1mi with 5280 ft., also line up 60 min. with 1 hr. and 60 sec with one minute. Then multiply the top and bottom and divede and you will have 46 mph. I threw 46 mph.

Doing the Math (Colin)

For doing Colin's throw we needed to see his data:

About Colin: Colin is 5'7 and is weighing in at 116.4 lb. He runs cross country in the fall for the LHS Cross Country team. In the spring Colin play baseball for the LHS Bolts (Junior Varsity). Colin also likes to participate in Martial Arts. This is how Colin's throw went.

This is how Colin is how Colin found out how fast his throw in miles per hour:

This is how Colin did his work:

Vertical Velocity: Like explained beforehand the equation for our Vertical Velocity is ax^2+bx+c. In my equation a equaled -16.1 because it is half of our negative acceleration to gravity which was 32.2. My b is the Velocity we are looking for and my c was 5.5 because that was the height in feet of my arm when I threw the ball. This made my equation look like, y = -16.1x^2 + bx + 5.5. Then to solve this I substituted x and y for a time and height during my throw. The x and y I substituted was y = 0 because that was the height at time x. X was represented by the amount of time the ball was in the air so x = 2.27. This made my equation 0 = -16.1(2.27)^2 + 2.27b + 5.5. Then I simplified the -16.1(2.27)^2 into -16.1(5.1529) because 2.27^2 is 5.1529. My equation was very simple after that. I multiplied the -16.1(5.1529) into -82.96. After that I added 82.96 so the equation was 82.96 = 2.27b + 5.5. At the same time I had subtracted 5.5 from both times to get the b alone. This created the equation 77.46 = 2.27b. I then divided both sides by 2.27 to get a single b and find the end vertical velocity of 34.123 = b.

Horizontal Velocity: To find the horizontal velocity of my throw was relatively simple. To do this I found how far I threw the ball, 52 yards, and multiplied it by 3 to get it into feet. This became 156 feet. Then I divided 156 by 2.27 because it was my total carry time. This showed me that the Horizontal Velocity was 68.72

Total Velocity: To find my total velocity I had to combine both my horizontal and vertical velocity into a pythagorean theorem equation. This equation looked like 34.123^2 + 68.72^2 = c^2. My Vertical velocity was a and my horizontal was b. I simplified the 34.123^2 and the 68.72^2 and got 1164.38 and 4722.4384. Then I added them together to get 5886.81. After that I square rooted both sides to get rid of the c^2. This gave me my total velocity answer of 76.72 = c.

MPH: To transfer my total velocity from feet per second to miles per hour I created the graph shown above. To create that graph I put feet over second for feet per second. Next to that I put 1 over 5280 for how many feet are in a mile. Then I put how many seconds are in a minute and how many minutes are in an hour. I then multiplied across each line, 76.12 * 1 * 60 * 60 =276,192. 1 * 5280 * 1 * 1 = 5,280. Then I divided 276,192 by 5,280 because it was in a fraction. This gave me my final answer of 52.309 MPH.

Analysis of data

The resluts are in. Colin threw the fastest out of all of us. With a whopping 52 mph he won the competition and smoke me and Andre by atleast 6 mph alltogether. Finding out who threw the fatsest was just the first part of our experiment. We also needed to answer four wrap up questions that were given to us from our math teacher.

"I can't throw the fastest. Actually, I threw the slowest" - Seth

Question #1: Which ball had more initial total velocity? For the question we have Colin winning that with a total initial velocity of 76.72 fps.

Question #2: How high did each ball go? For this problem we had to do some work. Our work for solving this problem is down below.

I took the equation h = -b/2a, and I substituted b for vertical velocity, and a for -16.1 which was the half of acceleration due to gravity. Then I solved this for everyone to by substitiuting this answer in for x in the equation y = -16.1x^2 + bx + c. Then I found that Colin threw the highest.

Question #3: Does more height automatically mean more initial total velocity? In our case, yes. Because of the data we can see that the one with the most height means more initial velocity. This is true because part of our calculations for total velocity is using vertical velocity which ahs to do with how far you threw it.

This is the graph of height and time.

Question #4: Does more horizontal distance mean more initial total velocity? In this case, yes, Colin had more horizontal distance and had the highest initial total velocity. Andre had the second farthest throw and had the second highest initial total velocity. Also, Seth had the shortest throw and the least amount of initial total velocity.

The End

Through this contest we found out that Colin threw the fastest and won. This could have been more easy with a radar gun......

Credits:

Created with images by Wokandapix - "baseball summer game" • Wokandapix - "calculator math mathematics" • kaboompics - "calculator scientific numbers" • Meditations - "algebra analyse architect" • TeroVesalainen - "question mark why problem" • scragz - "End"

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