Tennis Point Lab Report
To conduct our experiment, we used a meter stick on concrete and a tennis ball. We dropped the ball by intervals of ten from 40-100 centimeters above the concrete. To get the average ball-drop height, we dropped the ball three times from each height, found the mean. Our Y value represents the average bounce heights in centimeters and X represents the drop height in centimeters. We created a scatter plot to show our data (below).
The line of best fit went through the points (80,48.3) and (90,54.3). To find the slope of the line we used the slope formula: M=(Y2 -- Y1)/(X2 -- X1). If (80,48.3) is X1 and Y1 then (90,54.3) is X2 and Y2. After this you take X2 and subtract it by X1 and the same with Y1 and Y2. So 90-80 is 10 and 54.3 48.3 is 6. …show more content…
Plugging it into the formula you should get m=6/10 which equals 0.6. We use this number to plug into Y=MX+B. To find B we need to find an X and a Y. Your X and Y can be any of the points your line goes through. We chose (90,54.3) and plugged it into the equation which looked like this 54.3=0.6(90)+B. First multiply 0.6 and 90 and that equals 54. Now we have 54.3=54+B, to find be just subtract your X value from both sides. B in the end equals (B=0.3).
To find our least square regression line we used a graphing calculator and our data. The formula for least square regression line is Y=AX+B. On the graphing calculator press STAT. Then from the menu press edit, then enter. We then entered all our X values in the column that says L1 and all Y values in L2. Press STAT again and move the cursor over CALC then we chose option #4 LinReg (AX+B) and press enter. For A it gave us .6 and for B it was 3.3. When plugging that into Y=AX+B you get Y=.6X+3.3. The least square regression line is an exact measurement whereas the line of best fit is what you can see. The best one to use is the least square regression line because it gives us a precise measurement rather than an educated guess.
With all of our data calculated we can now predict what would happen if we changed the X or Y variable. For X we increased the bounce height to 3 meters
(300 cm) and calculated it both through the line of best fit and least square regression line.
For the least square regression line all you have to do is plug 300cm into the X spot in the equation. Our equation Y=.6X+3.3 will now look like Y=.6(300)+3.3. We first multiplied .6 and 300 because of order of operations. Then we added 3.3 and the bounce height was 183.3. Using the line of best fit we first used the formula Y=MX+B and plugged in the M and B we found which looked like this y=0.6X+0.3. Then for X we put in 300, Y=0.6(300)+0.3. Just like the last equation we multiply .6 and 300, then added .3 so we got 180.3 as the bounce height for slope intercept. Both the predictions are so close that you could choose either equations because they are both extremely close to each …show more content…
other.
Next we predicted the drop height, if the bounce height was 2 meters or 200cm.
We chose to use the least square regression line because it is more accurate than the slope intercept line, because the calculator takes our data and gives us an exact number whereas the slope intercept form is an educated guess we make with our eyes. We use the same formula as before Y=.6X+3.3 and instead of plugging in X we plugged in Y. We did this because we need to use the exact formula as before in order to find our prediction. Our equation now looks like 200=.6X+3.3. We need to get X alone so we subtracted 3.3 from 200 which is 196.7. Then we divided .6 from 196.7 and we got 327.8. To check our math we put 327.8 back into our last equation Y=.6(327.8)+3.3. and we got 199.98 which rounded up is
200.
If the drop height was 0cm we would get a bounce height of 0cm because it is impossible to get an outcome out of nothing. The point (0,0) is the origin on the graph and both of our lines don’t go through the origin point. This could be because of human error or slight differences in the way we dropped it or how the tennis ball hit the pavement. Our data isn’t as exact because we didn’t do it in a controlled environment and there were some other factors that would have caused our line to not go through the origin.
The line of best fit went through the points (80,48.3) and (90,54.3). To find the slope of the line we used the slope formula: M=(Y2 -- Y1)/(X2 -- X1). If (80,48.3) is X1 and Y1 then (90,54.3) is X2 and Y2. After this you take X2 and subtract it by X1 and the same with Y1 and Y2. So 90-80 is 10 and 54.3 48.3 is 6. …show more content…
Plugging it into the formula you should get m=6/10 which equals 0.6. We use this number to plug into Y=MX+B. To find B we need to find an X and a Y. Your X and Y can be any of the points your line goes through. We chose (90,54.3) and plugged it into the equation which looked like this 54.3=0.6(90)+B. First multiply 0.6 and 90 and that equals 54. Now we have 54.3=54+B, to find be just subtract your X value from both sides. B in the end equals (B=0.3).
To find our least square regression line we used a graphing calculator and our data. The formula for least square regression line is Y=AX+B. On the graphing calculator press STAT. Then from the menu press edit, then enter. We then entered all our X values in the column that says L1 and all Y values in L2. Press STAT again and move the cursor over CALC then we chose option #4 LinReg (AX+B) and press enter. For A it gave us .6 and for B it was 3.3. When plugging that into Y=AX+B you get Y=.6X+3.3. The least square regression line is an exact measurement whereas the line of best fit is what you can see. The best one to use is the least square regression line because it gives us a precise measurement rather than an educated guess.
With all of our data calculated we can now predict what would happen if we changed the X or Y variable. For X we increased the bounce height to 3 meters
(300 cm) and calculated it both through the line of best fit and least square regression line.
For the least square regression line all you have to do is plug 300cm into the X spot in the equation. Our equation Y=.6X+3.3 will now look like Y=.6(300)+3.3. We first multiplied .6 and 300 because of order of operations. Then we added 3.3 and the bounce height was 183.3. Using the line of best fit we first used the formula Y=MX+B and plugged in the M and B we found which looked like this y=0.6X+0.3. Then for X we put in 300, Y=0.6(300)+0.3. Just like the last equation we multiply .6 and 300, then added .3 so we got 180.3 as the bounce height for slope intercept. Both the predictions are so close that you could choose either equations because they are both extremely close to each …show more content…
other.
Next we predicted the drop height, if the bounce height was 2 meters or 200cm.
We chose to use the least square regression line because it is more accurate than the slope intercept line, because the calculator takes our data and gives us an exact number whereas the slope intercept form is an educated guess we make with our eyes. We use the same formula as before Y=.6X+3.3 and instead of plugging in X we plugged in Y. We did this because we need to use the exact formula as before in order to find our prediction. Our equation now looks like 200=.6X+3.3. We need to get X alone so we subtracted 3.3 from 200 which is 196.7. Then we divided .6 from 196.7 and we got 327.8. To check our math we put 327.8 back into our last equation Y=.6(327.8)+3.3. and we got 199.98 which rounded up is
200.
If the drop height was 0cm we would get a bounce height of 0cm because it is impossible to get an outcome out of nothing. The point (0,0) is the origin on the graph and both of our lines don’t go through the origin point. This could be because of human error or slight differences in the way we dropped it or how the tennis ball hit the pavement. Our data isn’t as exact because we didn’t do it in a controlled environment and there were some other factors that would have caused our line to not go through the origin.