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I am going to use some physics principals to find out the height of the shot; the range of the shot and the initial velocity of the ball

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Introduction

Planning Introduction The experiment was that two video cameras filmed a tennis launcher firing a tennis ball into the sky. One camera was set up near the launcher to get the 7 slides of the ball's motion. Each slide is separated by 0.04 seconds. It was pointing at right angles to the path of the ball. The experiment was repeated with another camera set up far away the launcher to get 25 slides of the ball's motion. Each slide is separated by 0.08 seconds. Aim: I am going to use some physics principals to find out the height of the shot; the range of the shot and the initial velocity of the ball. I will plot the graphs to find patterns of the data then associate the graphs to find out some information. Also I will compare the value of near shot with the value of far shot. Theory: Formulae: a = (v - u) / t t = (v - u) / a v = u + at v2 = u2 + 2as s = ut + 1/2 at2 "s"--Displacement ( metre) "t"--Time ( second) "u"--Initial Velocity ( metre/second) "v"--Final Velocity ( metre/second) "a"--Acceleration ( metre/second2) Prediction These two graphs are drown roughly. ...read more.

Middle

Range Secondly I want to calculate the range of the near shot. Form the graph you can see the distance in an interval time does not change too much, because the vertical acceleration of the ball is 0 if ignore the air resistance. Without the acceleration the velocity will not change at all. In the real life you can't ignore the air resistance so the rate of increasing the distance is slight changed. First I should work out the initial velocity horizontally. Basically the method is the same as calculating the initial vertical velocity. "u" - initial horizontal velocity "D1" - distance of the first point "D2" - distance of the second point "t1" - time of the first point. "t2" - time of the second point u = (D2 - D1)/ (t2-t1) = (0.54 - 0.00)/ (0.04-0.00) = 13.50m/s The time for the ball to reach the highest point is the same as to move the half range so if I can work out how long it takes to reach the highest point then I can calculate the range. Vertical: a = (v-u)/t t = (v - u)/a = (0 - 9.50) / (-9.80) = 0.97s Horizontal: s = ut = 13.50 � 0.95 = 13.10cm = 13.10m So the whole range = 13.10 � 2 = 26.20 m Here you should separate the calculation into two parts vertical and horizontal. ...read more.

Conclusion

The uncertainty of taking the value of the time is+/-0.002s There are some errors on the basic data. I have labeled on the graph. Data Near Shot Displacement/cm Photo Scaled up Photo Time/s x y x y 1 0.00 0.00 0.00 0.00 0.00 2 0.04 2.70 1.90 54.0 38.0 3 0.08 5.50 3.50 110 70.0 4 0.12 8.00 5.00 160 100 5 0.16 10.80 6.90 216 138 6 0.20 12.50 7.50 250 150 7 0.24 15.10 8.90 302 178 Far Shot Displacement/cm Photo Scaled up Photo Time/s x y x y 1 0.00 0.20 0.10 22.8 11.4 3 0.08 1.50 0.90 171 103 5 0.16 2.70 1.80 308 205 7 0.24 4.00 2.50 456 285 9 0.32 7.20 3.00 821 342 11 0.40 6.40 3.50 730 399 13 0.48 7.50 3.90 855 445 15 0.56 8.50 4.30 969 490 17 0.64 9.60 4.50 1094 513 19 0.72 10.6 4.60 1208 524 21 0.80 11.6 4.50 1322 513 23 0.88 12.7 4.40 1448 502 25 0.96 13.7 4.20 1562 479 27 1.04 14.7 3.90 1676 445 29 1.12 15.8 3.60 1801 410 31 1.20 16.6 3.20 1892 557 33 1.28 17.5 2.90 1995 331 35 1.36 18.5 2.40 2109 274 37 1.44 19.3 1.70 2200 194 39 1.52 20.1 0.90 2291 103 41 1.60 21.0 0.40 2394 45.6 43 1.68 21.7 -0.70 2474 -79.8 45 1.76 22.6 -1.30 2576 -148 ...read more.

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