Vertical Motion Investigation

In math we got assigned to throw a ball back and forth. It was not just throwing it for a random reason. We have to find out various things; the x intercept, y intercept the vertex, the initial height, the vertical motion and the highest height it attained.

In groups of 4 we went out and found a spot to do so. We placed a meter stick as vertical as possible for the measurement. We then recorded Ali and I throw a tennis ball back and forth. We had some problems with finding where to do it. As in some places such as against the wall inside the building the meter stick would not stay up as straight as possible. In other places people kept on interrupting is. We also had the problem at our last spot however we chose to ignore it and stay there since everything else was perfect. We tried to throw the ball a couple of times. we then realized in order to get the most accurate result we must stand as close to the meter stick as possible. We then proceeded with filming and recorded one video of us throwing it back and forth a couple of times.

It would not let me insert the video, because I do not have premium.

Throwing the ball video

Afterwards we put the video into tracker without any problems. We did have some problems downloading tracker. Since I was the only one who already half downloaded it, for the others it did not work. We tried various things on mine but it would not work. However at the end of class it finally worked and we got it installed. Anyways when we inserted the video into tracker it was already the right way up. We then had to had to choose which throw we wanted to use; select the frame of interest. We looked at all the throws and the throw we selected was thrown the closes to the meter stick height and was the best throw.  We started the frame right after it left my hand and the end of the frame was right before it came into Ali’s hand. The next step we had to do to proceed our investigation was to calibrate the video. We inserted a calibration stick. We had to do this so there is a measurement.  Our calibration stick was 100 centimeters. We then pressed the button next to it to show the coordinate axis. We then dragged the center of the axis on the position where it has left my hand. We then pressed on the axis button again. We then had to create point mass to plot the data. We tracked the ball throughout the throw. We held down the shift key and pressed on the ball at every point. Afterwards we then proceeded with changing the y axis so our data is distance over time.


Screen Shot 2016-11-23 at 09.32.34.png
This is the video in tracker with the points of the ball.


After we changed the y axis.







We then copied our time data and y data over into excel. In excel we had to change the data into numbers before we could transfer it onto desmos. When we put it into desmos we changed the x axis so the parabola was easier to see. We then proceeded which trying several formulas to find out the y intercept. All of them did not work. Afterwards we realized all the formulas were in a video on schoology. The formula we used to get the y axis was  y1 ~ ax^21 + bx1 + c. The y intercept was -105.29. We continued on finding the vertex where the formula was y1 ~ a(x1 – h)^2  + k. The vertex was 0.52196, 44.427. We then proceeded on finding out the x intercept. We used the formula to find out the y intercept so the line of best fit would appear. We then pressed on where the line met the 0 and we got our data. Our x intercepts were 0.2376 ,0 and 0.8063, 0. The highest point that the ball travelled was 150 centimeters above the ground. The initial velocity of the ball was 7.72 m/s . When the height is 0.27 the velocity was 7.72 m/s. How we calculated it was 0.27 was the first data point when the ball left the hand. The initial velocity was given to us on tracker therefore we did not need to find it. The speed of the ball is 7.72 m/s.

This is our data in desmos with the x axis and the way we found the y axis.


The reason I know our solution is incorrect is by using the vertical motion formula as well as our quadratic formula solution: H(t)=-5t^2 + VoT + Ho. In order to get our answer I first have to substitute t in for 0. Look at the first post-it note for my calculation. However we had problems doing our last calculation, because the throw we chose was a negative throw meaning it went back. Therefore it was a negative in the square root. It would not let us do

the calculation.   We had to change our calculation to make it positive.753a265a-bf18-4831-964e-3352169e7075

Our solution to the problem is this. This is in the quadratic formula. The a is suppose to be as close to -5 as possible since it is -5t. However ours is -549. When we substitute in the values the answer us y = -549^2 + 573 – 105. However our answer is wrong, because our throw went in the wrong direction. Therefore when we substitute in the numbers it is unfactorizable. As well as the constant for gravity is wrong.  screen-shot-2016-11-28-at-10-17-59

The real world application is that the quadratic formation we did is accurate, because if we have a program which calculates it how different is it to how we would calculate it. There are several real life aspects which change the throw; air resistant and the rotation of the ball.




Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s