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Data Analysis - Two cameras were set up to record images of a tennis ball launcher firing a tennis ball into the sky.

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Data Analysis Coursework Introduction Two cameras were set up to record images of a tennis ball launcher firing a tennis ball into the sky. Camera 1 was set up in line with the camera launcher and filmed the beginning of the launch. The film was then digitised and 7 sides were produced of the ball's motion. Each slide is separated by 0.04 seconds (these slides are numbered from 1 to 7). Camera 2 was set up in line with what was thought to be the halfway point of the ball's motion. This camera filmed the whole of the ball's motion from the start to the end of it. The film was then digitised into 25 slides of photos. Each slide is separated by 0.08 seconds (these slides are numbered from 1 to 45 in odd numbers). Planning I was given an excel file fill with some measurements from the physics department. My teacher showed me the slides, which are printed out. After I had a chance to see it I recognize that they were taken a few years ago and it took place at the tennis court outside the physics department. The measurements that I have got were the near shot displacement and the far shot displacement. They measure by measuring the horizontal and vertical distance of the tennis ball from the tennis launcher. ...read more.


By this stage we have got both the initial vertical velocity and initial horizontal velocity therefore we can plug in them into the equation with other measurements that we already know in order to work out the range of the ball. (UX=13.5m/s, UY=9.5m/s) As we already know that the final vertical velocity of the ball is 0m/s, we also know that the acceleration of the ball vertically is -9.8 as gravity is 9.8 m/s at this latitude so it is decelerating at 9.8 m/s. Plus there is no horizontal acceleration as there are no force acting upon as we are ignoring the air resistance. Here is how I am going to use these values in order to find out other results which I want by using the SUVAT equations: Vertical S= 4.8 U= 9.5 m/s V= 0 m/s A= 9.8 m/s T= 0.96s S= ut+1/2at� S= 13.5 x 0.96 + (0.5 x 0 x 0.96�) S= 12.96 x 2 2S= 25.92 m Maximum range is 25.92m Horizontal S= 25.92 U= 13.5 m/s V=? m/s A= 0 m/s T= 0.96s V= u + at v- u /a= t 9.5/9.8= 0.96s S= ut+1/2at� S= 9.5 x 0.96 + (0.5 x -9.8 x 0.96�) S= 4.6m Maximum height is 4.6m From these measurements, I have found out that the initial vertical displacement is 25.92m/s and the horizontal vertical displacement is 4.6m/s. ...read more.


S= 12.96 x 2 S= 25.92 m Maximum range is 25.92m (2592cm) Results from data is 2576cm (scaled up X) Horizontal S= 25.92 U= 13.5 m/s V=? m/s A= 0 m/s T= 0.96s V= u + at v- u /a= t 9.5/9.8= 0.96s S= ut+1/2at� S= 9.5 x 0.96 + (0.5 x -9.8 x 0.96�) S= 4.60416m Maximum height is 4.6m (2.d.p) (460cm �0.5cm) Results from data is 524cm(scaled up Y) I believe that if we are not ignoring air resistance, the vertical distance shouldn't be that close with the data, so I think there was a bit of error with both of the distances, but overall the error was accepted. Expect air resistance, other factors that will effect the overall measurements are gravity, the errors when measuring the distance with the photos, and the angle of the camera when it is recording the movement of the ball. Because if the camera wasn't placed perpendicular with the line that the ball travelled, it will make the measurements go wrong and affects the results of the experiment. Because I didn't actually saw the experiment when it was taken place so I'm not sure about this problem. But overall the calculations and the results are very close together this proves that the method of which I used is right. Page 1 12/18/2007 ...read more.

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