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Lab Report 3 - the slope and the intercept(VI 5 Then I...

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= 105.2 cm = 29 ° = 37.7 cm # D (cm) d1 (cm) d2 (cm) d3 (cm) d4 (cm) d` (cm) H - d` (cm) h / D 1 74.8 26.0 26.1 26.3 25.4 26.0 79.3 1.059492 2 63.3 52.6 53.9 55.0 55.4 54.2 51.0 0.805292 3 55.3 70.5 70.6 71.5 73.1 71.4 33.8 0.610759 4 46.9 84.9 85.6 85.6 85.7 85.5 19.8 0.421109 5 36.7 96.8 98.1 98.0 98.6 97.9 7.3 0.199591 tan 29° 0 -0.6 EXPERIMENT M4 PROJECTILE MOTION S (VI 1) In this experiment, we dropped the ball along the apparatus from a constant height and let it hit the sensing paper on the board. We repeated the procedure 4 times each for 5 different distances from the board to the point of launch of the ball. (VI 2) I then plotted a graph h/D vs. D and included the value of tan θ as the intercept H θ y = 0.021x - 0.572 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 0 20 40 60 80 Experiment to determine launch speed Date Linear (Date)
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= cm/s^-1 = cm/s^-1 (i) (ii) b σ m b^2 + 1 2 * m = 57.04464 cm/s^-1 (VI 6) I then calculated the error in Vo using the possible errors in obtaining the values of
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Unformatted text preview: the slope and the intercept (VI 5) Then I calculated the value of v o using the slope of the graph The values obtained by the 2 methods are significantly different from each other. There may be a number of reasons for this including Air resistance Errors in obtaining the data, especially finding the correct point where the ball stuck the sensing paper = 369460 7 52780 v = 229.7389823 cm/s^-1 = 0.021 (from graph) (VI 4) Then I calculated the value of v o using the slope of the graph σv = v * ^ 2 + ^ 2 ^ 0.5 cm/s^-1 v ^2 = 10*g*S = 10*980*37.7 7 7 30827.49813 v = 175.5776128 = 980 2*0.021*(0.87^2) 980 0.0317898 = 2*(slope)*cos^2° g 2*(v ^2)*cos^2° Slope = g v ^2 = a sadf a gwaegaweg awegweg...
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