PHY-Lab5

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Camden County College *

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201

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Mechanical Engineering

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Dec 6, 2023

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docx

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Uploaded by DeanFangLoris38

Tu Pham Lab5: Horizontal Projectile Motion 1. Objective: to calculate final velocity and angle of approach at ground for given objects that follow projectile motion. 2. Tools: marble, steel, projectile launcher, 2m ruler, spirit level, carbon paper (transfer paper), white paper, paper tap 3. Procedure: 1) Setup and Marking: Use paper tape to mark the point where the projectile stands on the ground. 2) Setting the Height: Use the 2-meter ruler to set the height accurately. Ensure that the projectile launcher is set up level with the ground using the spirit level. 3) Test Shot: Conduct a test shot to place the paper setup. Put a piece of black paper over a white one and place it at the test point of collision. 4) Experiment Runs:  For each type of projectile (marble and steel), perform three runs at each of the selected heights. This is a total of 18 trials.  Before each run, set up the projectile launcher at the desired height, ensuring it's level.  Load the projectile into the launcher, aim it horizontally, and fire it.  Make sure to measure and record the distances from the points of collision to the marked point on the ground for each trial at every height. 4. Data: Object Heigh t h(m) Range x(m) x avg = x 1 + x 2 + x 3 3 v y = √ 2 gh (m/s) t = √ 2 h g (s) v x = x t (m/s) v = √ v x 2 + v y 2 (m/s) θ = tan − 1 | v y v x | ( ° ) Marbl e 1.0 x 1 = 0.863 x 2 = 0.963 x 3 = 0.750 x avg = 0.859 √ 2 × 9.8 × 1.0 = 4.427 √ 2 × 1.0 9.8 = 0.452 0.859 0.452 = 1.90 √ 1.90 2 + 4.427 2 = 4.82 tan − 1 | 4.427 1.90 | = 66 Steel 1.0 x 1 = 0.771 x 2 = 0.758 x 3 = 0.767 √ 2 × 9.8 × 1.0 = 4.427 √ 2 × 1.0 9.8 = 0.452 0.765 0.452 = 1.69 √ 1.69 2 + 4.427 2 = 4.74 tan − 1 | 4.427 1.69 | = 69

Tu Pham x avg = 0.765 Marbl e 1.5 x 1 = 1.061 x 2 = 1.112 x 3 = 1.095 x avg = 1.089 √ 2 × 9.8 × 1.5 = 5.422 √ 2 × 1.5 9.8 = 0.553 1.089 0.553 = 1.97 √ 1.97 2 + 5.422 2 = 5.77 tan − 1 | 5.422 1.97 | = 70 Steel 1.5 x 1 = 0.899 x 2 = 0.926 x 3 = 0.942 x avg = 0.922 √ 2 × 9.8 × 1.5 = 5.422 √ 2 × 1.0 9.8 = 0.553 0.922 0.553 = 1.67 √ 1.67 2 + 5.422 2 = 5.67 tan − 1 | 5.422 1.67 | = 72 Marbl e 2.0 x 1 = 1.217 x 2 = 1.211 x 3 = 1.282 x avg = 1.237 √ 2 × 9.8 × 2.0 = 6.261 √ 2 × 2.0 9.8 = 0.639 1.237 0.639 = 1.94 √ 1.94 2 + 6.261 2 = 6.55 tan − 1 | 6.261 1.94 | = 72 Steel 2.0 x 1 = 1.050 x 2 = 1.087 x 3 = 1.110 x avg = 1.082 √ 2 × 9.8 × 2.0 = 6.261 √ 2 × 2.0 9.8 = 0.639 1.082 0.639 = 1.69 √ 1.69 2 + 6.261 2 = 6.49 tan − 1 | 6.261 1.69 | = 74 5. Result: In the experiment, we observed that as the height of the projectile increased, the final velocity also increased. This is consistent with the basic principles of projectile motion. Additionally, the approach angle (θ) increases as the altitude increases. A higher launch altitude results in a steeper angle of descent upon impact. These observations highlight the direct relationship between altitude, terminal velocity, and angle in horizontal projectile motion. The marble always has a higher velocity than the steel bullet in this experiment, possibly due to environmental conditions, and friction, ... 6. Conclusion In summary, the goal of this experiment is to calculate the final velocity and angle of approach to the ground for objects following projectile motion. As projectile height increases, we observe a corresponding increase in both the terminal velocity and the angle of approach (θ). This emphasizes the direct relationship between height, velocity, and angle in horizontal projectile motion. Furthermore, the velocity of the marbles is always higher

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