Lab Report #2

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Apr 3, 2024

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Report for Experiment #2 One and Two Dimensional Motion Serena Le Lab Partner : Ankee Zhang TA : Jinzheng Li January 23, 2024 Abstract In this experiment, we observed the motion and acceleration of an air hockey puck on a slanted surface in one and two dimensions to find the acceleration due to gravity, − g , in free fall. Investigation 1 focused on the setup and calibration of the webcam software used to record data and found an error of δx = 0.000381 cm and δy = 0.000403 cm in the position of a stationary puck, which increases by an order magnitude when in the following investigations. Investigation 2 analyzed the motion in one dimension along the incline, calculating accelerations of a y , 1 =− 51.437 cm / s 2 ± 0.061 , a y , 2 =− 55.830 cm / s 2 ± 0.050 , and a y , 3 =− 54.281 cm / s 2 ± 0.098 . Investigation 3 observed motion in two dimensions, across and along the incline, yielding accelerations a x =− 1.124 cm / s 2 ± 0.038 and a y =− 56.395 cm / s 2 ± 0.040 . The accelerations obtained in Investigations 2 and 3 were used in Investigation 4 to calculate the experimental values for the

acceleration due to gravity, where we found us a y =− g = 10.4 m / s 2 , which represents the scientifically accepted value of − g = 9.81 m / s 2 . Introduction In this lab, we replicated the experiments of Galileo using more modern technology and techniques to observe motion. Our objective was to observe the motion of an air hockey puck on a slanted surface in one and two dimensions, calculate the velocity and acceleration in one and two dimensions, and use the data to calculate an experimental value for the acceleration due to gravity. In our calculations, we used the known relationships for constant velocity and constant acceleration, broken down into its respective components: V x = ΔX ∆t ∧ a x = ΔV x ∆t for the x direction, V y = ΔY ∆t ∧ a y = ΔV y ∆t for the y direction. (I1.1) (I1.2) Throughout Investigations 2 and 3, we recorded data using a live camera software that tracked the position of a battery powered hover puck’s positions for each frame over small increments in time. Our goal was to record the motion in one and two dimensions and observe the pucks velocity in both dimensions. Since the camera only recorded position and not velocity, we had to calculate the average velocity of the puck between each time interval using the equation: V avg = ΔX Δt = X 2 − X 1 t 2 − t 1 at timet avg = t 2 − t 1 2 (I2.1) (I2.2) In Investigation 4, we applied our calculations of velocity and acceleration to forces and free fall. Our goal was to use our data from the previous investigations to find the puck’s acceleration as well as the acceleration due to gravity. In free fall, we recognize gravity as the only force acting on the particle in motion. The equation for gravity can be given as: F g = mg (I3.1) and can be applied to the equation for Newton’s second law to derive acceleration due to gravity in free fall: Σ F y = ma y =− mg a y =− g (I3.2) (I3.3) However, in the case of an object on an incline, the acceleration due to gravity can be calculated by: a y =− g sin θ, (I3.4) where θ is the angle of the incline above the x axis. Throughout this experiment, especially with the various calculations, we encountered various forms of error. To quantify these errors, we used standard deviation: σ = √ ( x ¿¿ 1 − x ) 2 +( x ¿¿ 2 − x ) 2 + … +( x ¿¿ n − x ) 2 n ¿¿¿ (I4) where x is the average of the given quantity, and standard error in the mean: δ x = σ √ n . (I5)

We also used different error propagation equations to calculate error propagation throughout different calculations, particularly for velocity, as given by: δV = V x √ ( δ ∆ X ∆ X ) 2 + ( δ ∆t ∆t ) 2 . (I6) Investigation 1 – Setup, Calibration, and Error Analysis Throughout this experiment, we used a three-legged glass table and a webcam with position- tracking software to record our measurements and data. Investigation 1 mainly focused on setting up and calibrating this equipment and quantifying the error from the setup. The first step was to level the table. Orienting the table so that the side with one leg was facing away from us, we turned the hockey puck on and placed it in the center of the table and adjusted the height of the table’s legs until the puck was near stationary. A wooden block was then placed beneath the singular leg in the back to create an inclined surface. We turned the puck off and opened the camera software to adjust the webcam until the entire table and wooden frame were in view and everything was in focus. Since the camera detects motion using color, under Camera Settings, we increased the exposure, brightness, saturation, and contrast to make the puck as bright as possible while keeping the frames per second at 30. We then mapped the table’s surface by selecting the outer corners of the table in the view settings. The sides of the table were then measured and input into the software accordingly. During this time, we also used a protractor to measure the slope of the incline, θ = 3 ° . Under Color Masking, we adjusted the settings until the puck appeared white and everything in the background appeared black while the puck was both stationary and moving. Once these adjustments were made, we turned the puck off, placed it in the center of the table, and clicked record experiment. We let the camera record the stationary puck’s position for about 10 seconds and pasted the data into Excel. Using Excel, we compiled the data into two histograms to display the frequency of the puck’s x and y positions, as shown in Fig 1.1 and 1.2, respectively. 33.473 33.481 33.489 33.497 33.505 33.513 0 10 20 30 40 50 60 70 80 90 Frequency of X-Position X-Position (cm) Frequency 32. 252 32.2 57 32. 262 32.2 67 32. 272 32.2 77 32. 281 32.2 86 32. 291 32.2 96 32. 301 Mo re 0 20 40 60 80 100 120 Frequency of Y-Position Y-Postion (cm) Frequency

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Using Eq (I4) and Eq (I5), we calculated the standard deviation and standard error on the mean of the x and y positions, as represented in Table 1 below. Tab1e 1 Standard deviation and standard error of the mean of the x and y positions of the puck. σx σy δx δy 0.006588 0.006965 0.000381 0.000403 However, these standard errors δx and δy only represent the uncertainty of the stationary pucks position. Thus, we increased the error by an order of magnitude so that the uncertainties of a puck’s position while in motion would be δx = 0.00381 cm and δy = 0.00403 cm in the x and y direction, respectively. These uncertainties were used to calculate propagated errors in Investigations 2 and 3. The main purpose of this investigation was to reduce the systematic error by leveling the table and quantify the instrumental error by recording the random fluctuations in the puck’s position as recorded by the camera. Though we did attempt to eliminate as much systematic and random error possible when leveling the table, there is always some inherent, unquantifiable amount of error that that exists and propagates throughout measurement. A slight tilt in the table to the left or right, for example, could affect the pucks motion and could introduce a bias—an acceleration—where there should not be one. As for instrumental error, we concluded a standard error in the pucks stationary position to be δx = 0.000381 cm and δy = 0.000403 cm and an uncertainty in the pucks position in motion to be δx = 0.000381 cm and δy = 0.00403 cm . Investigation 2 – One Dimensional Motion In Investigation 2, we observed the motion of the puck in one dimension—the y dimension, for the purpose of labeling in this experiment—to find the acceleration along the incline. The same inclined setup used in Investigation 1 was also used in this investigation as well as in Investigation 3. For data collection, the puck was turned on and placed towards the top. Still holding the puck at the top, my lab partner started the recording, and after about one second, I let go of the puck. We let the puck move down the incline and bounce a few times at the bottom, stopping the recording after about 8 seconds. The data was copied and pasted into Excel. To find the velocity of the puck throughout its movement, we used Eq (I2.1) and Eq (I2.2) to find the average velocity, v y ,avg of the puck at average time, t avg . An example of data in Excel is shown in Table 2 below. Since we are only observing motion in one dimension, data for x was disregarded. Table 2: Example of data table for the position and average velocity of the puck. t (s) y (cm) t avg (s) v y,avg (cm/s) 1.766 6 59.08 5 1.78325 1.78325 1.799 9 59.11 1.8166 1.8166

1.833 3 59.08 2 1.84995 1.84995 1.866 6 59.02 1.88325 1.88325 1.899 9 58.87 4 1.9166 1.9166 We then began calculating error. We used the uncertainty of a puck in motion from Investigation 1 to calculate the error in displacement, δ ∆Y , using the equation: δ ∆Y = √ δx 2 2 + δx 1 2 = √ 2 δy , (2.1) where δy = 0.000403 cm . This uncertainty was then applied to Eq (I6). We assumed the error in time to be negligible and simplified the equation to: δ V y = V y x √ ( δ ∆Y ∆ X ) 2 . (2.2) This equation was then applied to find the error in each of the v y ,avg calculations. The data was then separated into three segments by the different drops—initial drop, first bounce, and second bounce—and plotted in a scatter plot as separate series, each fit with a trendline and equation. Because the puck was dropped a second after the recording was started, the first ~1.75s of data were omitted from the first series. The plot is shown in Figure 2 below. 3 3.5 4 4.5 5 5.5 6 6.5 -60 -40 -20 0 20 40 60 f(x) = − 52.6 x + 286.7 R² = 0.99 f(x) = − 55.83 x + 226.96 R² = 0.98 f(x) = NaN x + NaN R² = 0 Vy (cm/s) vs. Time (s) Time (s) Veritcal Velocity (cm/s) Figure 2: Scatter plot of velocity in the y direction as a function of time, with the respective lines of best fit and their equations. We found the slope of the first drop, first bounce, and second bounce to be -51.438, -55.826, and -52.602, respectively. This slope represents the acceleration along the y-surface of the incline, a y , for this investigation. The data was also plotted in the IPL Straight Line Fit Calculator. The IPL plot is more accurate and precise because it accounts for the error of each measurement as well as the uncertainty in slopes. Here, we found the slopes, denoted as accelerations, and their respective

uncertainties to be a y , 1 =− 51.437 ± 0.061 cm / s 2 , a y , 2 =− 55.830 ± 0.050 cm / s 2 , and a y , 3 =− 54.281 ± 0.098 cm / s 2 . In conclusion, the acceleration of the puck along the incline was very similar for each drop segment, with the puck speeding up or slowing down at similar rates. Using the more accurate calculations from the IPL calculator, we found the accelerations of each drop to be a y , 1 =− 51.437 cm / s 2 , a y , 2 =− 55.830 cm / s 2 , and a y , 3 =− 54.281 cm / s 2 with an error of ± 0.061 , ± 0.050 , and ± 0.098 respectively. This data models a particle in constant acceleration, and the accelerations calculated are used later in Investigation 4. Investigation 3: Two-Dimensional Motion In Investigation 2, we observed the motion of the puck in two dimensions—in the y-direction along the incline and the x-direction across the incline—to find and compare the acceleration of the puck in both dimensions. As aforementioned, the setup remained the same as in Investigations 1 and 2. This time, instead of releasing the puck from rest, we pushed the puck with an initial velocity from the bottom left-hand corner of the puck. We started the recording and let it run for about 1 second before pushing the puck at an angle so that it had an initial velocity in both the x and y directions and created a parabolic shape along the table. It took multiple attempts to try and get the puck to reach the top edge of the table before landing in the bottom right-hand corner. We made sure to stop each recording right before the puck hit the corner. We took the data from our best attempt and pasted it into Excel. Similar to Investigation 2, we used Eq (I2.1) and Eq (I2.2) to find the average velocities, v x ,avg and v y ,avg of the puck at average time, t avg . An example of data in Excel is shown in Table 3 below. Table 3: Example of data table for the position and average velocities in the x and y directions of the puck. t (s) X (cm) Y (cm) t avg (s) v x,avg (cm/s) v ,y,avg (cm/s) 1.6335 7.188 8.695 1.6501 27.62048 61.86747 1.6667 8.105 10.749 1.68345 25.8806 54.80597 1.7002 8.972 12.585 1.7168 28.01205 58.58434 1.7334 9.902 14.53 1.7501 27.87425 55.1497 1.7668 10.833 16.372 1.7834 26.20482 48.55422 1.8 11.703 17.984 1.81675 28.23881 50 1.8335 12.649 19.659 1.8501 26.53614 45.72289 1.8667 13.53 21.177 1.8834 27.87425 45.08982 The same methods to calculate error in Investigation 2 were applied to this data as well. This time, we used both δx = 0.000381 cm and δy = 0.00403 cm from Investigation 1 in Eq (2.1) to find the uncertainties in all of the x and y velocities, δ V x and δ V y , respectively. The data was then compiled in to two scatter plots, based on their respective directions, each fit with a trendline and equation. Because the puck was dropped a second after the recording was started, the first ~1.62s

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of data were omitted from each graph. The plots of v x and v y as functions of time are shown in Figure 3.1 and Figure 3.2, respectively, below. 1.5 2 2.5 3 3.5 4 -10 -5 0 5 10 15 20 25 30 35 f(x) = − 2.34 x + 32.05 Vx (cm/s) vs Time (s) Time (s) Horizontal Velocity (cm/s) Figure 3.1: Scatter plot of velocity in the x direction as a function of time, with the line of best fit and its equation. 1.5 2 2.5 3 3.5 4 -60 -40 -20 0 20 40 60 80 f(x) = − 54.84 x + 144.81 Vy (cm/s) vs Time (s) Time (s) Vertical Velocity (cm/s) Figure 3.2: Scatter plot of velocity in the y direction as a function of time, with line of best fit and its equation. Again, the slope of these graphs represents the accelerations in the respective directions. The slope of Figure 3.1 was -2.342, representing an acceleration a x =− 2.342 cm / s 2 in the x-direction, or across the incline. The slope of figure 3.2 was -54.843, representing an acceleration a y =− 54.843 cm / s 2 in the y-direction, or along the incline. The data was also plotted in the IPL Straight Line Fit Calculator to factor in uncertainty and determine more accurate accelerations. Here, we found the accelerations and their respective uncertainties to be a x =− 1.124 ± 0.038 cm / s 2 and a y =− 56.395 ± 0.040 cm / s 2 . In conclusion, the acceleration of the puck along the incline was very similar to those in Investigation 2, with the puck speeding up or slowing down at similar rates. On the other hand, the acceleration of the

puck across the inline was closer to zero, which is the theoretical value of acceleration in the x- direction in parabolic/free fall motion. Using the more accurate calculations from the IPL calculator, we found the acceleration across the incline to be a x =− 1.124 cm / s 2 with an error of ± 0.038 and the acceleration along incline to be a y =− 56.395 cm s 2 with an error of ± 0.040 . This data is similar to that of a particle in free fall, and the accelerations calculated are used in Investigation 4. Investigation 4 – Acceleration Due to Gravity This investigation did not require any setup, but rather focused on the analysis of data gathered in Investigations 2 and 3. The purpose of this investigation was to use the accelerations calculated in Investigations 2 and 3 to determine an experimental value for the acceleration due to gravity, a y =− g , or the acceleration of a particle in free fall. The derivation of this acceleration is included in the Introduction in Eq (I3.1-I3.3). To do this, we first drew a free body diagram to identify the forces acting on the puck during Investigations 2 and 3. The drawing is shown in Figure 4.1 below. Figure 4.1: Free body diagram of the puck on the inclined table, where the surface of the table is labeled as ‖ and the perpendicular axis is labeled as ⊥ , along which the normal force, F N , lies . These axes are not to be confused with the x axis and y axis, along which the force of gravity, F g , lies. Using this, we broke the force of gravity down into components, F g,‖ = F g sin θ and F g, ⊥ = F g, ⊥ = F g cos θ , where θ is the angle of the slope, as determined to be θ = 3 ° in Investigation 1. Since the puck did not move above or below the surface, the perpendicular

acceleration a ⊥ = 0 . Applying the equation for gravity, Eq (I3.1), we get F g,‖ = mg sin θ . Plugging this into Newton’s second law, we get: Σ F ‖ = m a ‖ =− mg sin θ a ‖ =− g sin θ . (4.1) (4.2) Since we are trying to find the experimental value for a y =− g , we continue a step further to isolate g : g = − a ‖ sin θ . (4.3) With this derivation, we applied our acceleration from in Investigations 2 and 3. Formerly labeled as a y , 1 , a y , 2 , and a y , 3 in Investigation 2 and as a y in Investigation 3, these are substituted into Eq (4.3) as a ‖ to yield g 1 = 982.804 cm / s 2 , g 2 = 1066.758 cm / s 2 , g 3 = 1037.170 cm / s 2 , and g 4 = 1077.565 cm / s 2 . Converting to proper SI units, kg s 2 m , and adjusting signage for easier calculations, we get g 1 = 9.83 m / s 2 , g 2 = 10.7 m / s 2 , g 3 = 10.4 m / s 2 , and g 4 = 10.8 m / s 2 . All values for the following equations are listed in Table 4.1 below. Table 4.1: Data including acceleration along the incline as determined in Investigations 2 and 3, the angle of the incline, the value of g calculated using Eq. (4.3), and all necessary components used to compute each quantities’ respective errors, which are also listed. Trial 1 2 3 4 a ( cm s 2 ) -51.436 -55.8298 -54.2813 -56.3954 a ( m s 2 ) -0.51436 -0.5583 -0.54281 -0.56395 δa 0.061 0.050 0.098 0.042 δa / a - 0.1185 9 - 0.0895 6 - 0.1805 4 - 0.0744 7 θ ( rad ) 0.05236 0.05236 0.05236 0.05236 δθ 0.008727 0.008727 0.008727 0.008727 g ( m s 2 ) 9.828042 10.66758 10.3717 10.77565 δg 0.652989 0.535237 1.049065 0.449599 sinθ 0.0523 36 0.0523 36 0.0523 36 0.0523 36 cosθ 0.99863 0.99863 0.99863 0.99863 δsinθ 0.008715 0.008715 0.008715 0.008715 δsinθ / sinθ 0.1665 14 0.1665 14 0.1665 14 0.1665 14 δg 0.2044 3 0.1890 7 0.2456 06 0.1824 1 To find the error for each of calculation of g, we apply the error equation:

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δg = g √ ( δa a ) 2 +( δ sin θ sin θ ) 2 , (4.4) where, δa is the error in the slope of each plot as determined in Investigation 2 and δ sin θ is the error in the sine function, as given by : δ sin θ = δθ ¿ (4.5) Since θ was a directly measured quantity, we assumed the instrumental error δθ to be half the smallest increment on the protractor, 0.5°. Converting to radians and plugging into the equation, we get δ sin θ =( 0.5 π 180 ) ¿ (4.6) With this, we plug into Eq (4.4) with the respective δa , a , and g values to get δ g 1 = 0.204 , δ g 2 = 0.189 , δ g 3 = 0.246 , and δ g 4 = 0.182 . We also found the average value for g , g , using the equation: g = g 1 + g 2 + g 3 + g 4 4 . (4.7) to get g = 10.4 m / s 2 . Using Eq (I4) we get a standard deviation of σ = 0.367 which is plugged into Eq (I5) to compute a standard error in the mean, δ g = 0.212 (values in Table 4.2 below). Table 4.1: Average value for g, standard deviation, and standard error within the mean. g σ δ g 10.410 74 0.367487 0.212169 In conclusion, the accelerations calculated in Investigations 2 and 3 were used to compute the experimental values of g to get g 1 = 9.83 m / s 2 , g 2 = 10.7 m / s 2 , g 3 = 10.4 m / s 2 , and g 4 = 10.8 m / s 2 . Further, we were able to compute their respective errors propagated through calculations giving us g 1 = 9.83 m / s 2 ± 0.204 , g 2 = 10.7 m / s 2 ± 0.189 , g 3 = 10.4 m / s 2 ± 0.246 , and g 4 = 10.8 m / s 2 ± 0.182 . Averaging these values and computing the standard error, we get our “best value” g = 10.4 m / s 2 ± 0.212 . Since our original objective in this investigation was to find acceleration due to gravity, a y =− g , we get a y =− 10.4 m / s 2 ± 0.212 . Conclusion In Investigation 1, we focused on calibrating the measurement instruments to reduce as much instrumental and systematic error as possible and keep that error from propagating throughout calculations. The random fluctuations in the pucks reported stationary position according to the camera alluded to an error of δx = 0.000381 cm and δy = 0.000403 cm and an uncertainty in the pucks position in motion to be δx = 0.000381 cm and δy = 0.00403 cm . In Investigation 2, we observed the pucks motion in one dimension to determine the acceleration along the incline. Taking the data from the initial drops and subsequent two bounces

off the bottom, we calculated accelerations of a y , 1 =− 51.437 cm / s 2 ± 0.061 , a y , 2 =− 55.830 cm / s 2 ± 0.050 , and a y , 3 =− 54.281 cm / s 2 ± 0.098 . In Investigation 3 had a similar procedure, except instead of letting the puck fall from the top edge to the bottom in a straight line, the puck was pushed from the bottom left corner with an initial velocity above the x-axis and create a parabolic shape and land in the bottom right corner, giving it motion both along and across the incline. Doing this, we found the acceleration across the incline, a x =− 1.124 cm / s 2 ± 0.038 , and the acceleration along incline, a y =− 56.395 cm / s 2 ± 0.040 . This acceleration across the incline, -1.124 cm/s 2 , models the theoretical x-acceleration of a particle in free fall, were a x = 0. In Investigation 4, we used the 4 previously calculated y-accelerations to find the acceleration due to gravity, a y =− g . We found each experimental value for g, g 1 = 9.83 m / s 2 ± 0.204 , g 2 = 10.7 m / s 2 ± 0.189 , g 3 = 10.4 m / s 2 ± 0.246 , and g 4 = 10.8 m / s 2 ± 0.182 , as well as the “best” or average value, g = 10.4 m / s 2 ± 0.212 , all with their respective errors. This in turn gives us a y =− 10.4 m / s 2 , which represents the scientifically accepted value of − g = 9.81 m / s 2 . The main sources of error stem from the webcams software’s limit of precision, causing instrumental error, and the likely imperfect leveling of the table, causing systematic and random error. We tried to eliminate errors by accounting for it throughout our calculations, using error propagation, standard deviation, and standard error in the mean. Questions 1. Find the time for the puck to reach the bottom of the air table starting from the top for Investigations 2 and 3. Are these times all the same, as shown in the picture of the cannon? Why or why not? The time it takes for the puck to reach the bottom of the air table from the time can be found using the kinematic equation: ∆Y = v 0 , y t + 1 2 a y t 2 , where ∆Y =− 65 cm , v 0 , y = 0 , and a y is the slope of the graph from the investigation. Using the first slope from Investigation 2 and the slope from Investigation 3, we get: Investigation 2: − 65 = 1 2 ( − 51.44 ) t 2 t = 1.590 s

Investigation 3: − 65 = 1 2 ( − 58.83 ) t 2 t = 1.540 s These times are very close and should theoretically be the same. This is because they both travel the same distance, both have a y-velocity of 0 m/s at the top and experience the same y-acceleration along the incline. 2. If the acceleration of a puck falling down an incline of angle θ (where θ < 45 ° ) is a θ , what is a 2 θ ? The acceleration of a 2 θ would be 2 a θ . As we see in Eq (I3.4), acceleration is directly proportional to sin θ , and for any angle θ < 45 ° , 2sin θ = sin 2 θ . This would mean the acceleration of a puck along an incline 2 θ where θ < 45 ° is twice the acceleration of the same puck along incline θ , or 2 a θ . 3. You determined the acceleration from the slope of your plot of v vs. t. What is the meaning of the intercept of your line with the v-axis? Do you expect this value to be close to the origin? The y-intercept represents the initial velocity of the puck. For the y-velocity in Investigation 2, I would expect this value to be close to the origin because the puck is released from rest, meaning it has an initial y-velocity of 0 m/s. In Investigation 3, on the other hand, I would expect the y intercept of both the v X vs. t and v y vs. t graphs to be above zero, as the puck was given an initial velocity greater than zero in both directions. 4. If you performed Investigations 2 and 3 again with an incline of θ = 0 , what would the trajectories look like in each case? In the ideal case where θ is equal to exactly 0°, the trajectory in Investigation 2 would be very different, in that the puck would not move at all. This is because on a completely flat surface, all forces would be balanced meaning there is no acceleration in any direction and the puck would stay where it is. On the other hand, in Investigation 3 the puck would move in a straight line across the table along the angle in which it was pushed until it hits the other side, bounces off, and keeps bouncing off the sides. 5. In Investigation 3, you launched the puck so that it goes all the way up and comes back down in the right corner of the table. Since the table is a square, the puck goes up the same distance in the y direction as it travels in the x direction. Calculate the theoretical angle that is required for this. To find the launch angle, θ , we must use the kinematic equations, separated into components: ∆ X = v 0 , x t + 1 2 at 2 = v 0 cos θt . (Q5.1)

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∆Y = v 0 , y t + 1 2 at 2 = v 0 sin θt − 1 2 gt 2 . (Q5.2) ∆ X = ∆Y when ∆ X = R (the range) and ∆Y = h (maximum height) with a magnitude of 65 cm. Knowing this, we adjust the equations: ∆ X = R = 2 v 0 2 sin θ cos θ g (Q5.3) ∆Y = h = v 0 sin θt h − 1 2 gt h 2 . (Q5.4) We recognize that time t h at ∆Y = h can be expressed as t total 2 , where t total is derived by setting ∆Y = 0 in Eq (Q5.2): ∆Y = v 0 sin θt total − 1 2 gt total 2 = ¿ v 0 sin θt total = 1 2 gt total 2 = ¿ t total = 2 v 0 sin θ g . This gives us t h = v 0 sin θ g , which we can plug into Eq (Q5.4) to get: ∆Y = h = v 0 sin θ v 0 sin θ g − 1 2 g ¿ (Q5.5) Now, we can set the Eq (Q5.3) and Eq (Q5.5) equal to each other and solve for θ : ∆Y = v 0 2 sin 2 θ 2 g = ∆ X = 2 v 0 sin θ cos θ g = ¿ sin θ 2 = 2cos θ = ¿ sin θ cos θ = 4 ¿ > tan θ = 4 = ¿ θ = tan − 1 ( 4 ) θ = 75.96 °. After solving, we get θ = 75.96 °, meaning that in order for the puck to reach a maximum height equal to the horizontal range of its motion, it must be launched 76° above the horizontal axis. References [1] IPL Straight Line Fit Calculator, http://www.northeastern.edu/ipl/data-analysis/straight-line-fit/ .

Lab Report #2

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