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5. If you were to throw a large log over the edge of the Grand Canyon and it took 5.65 seconds to hit the ground, calculate the velocity of the log at impact in \( \mathrm{m} / \mathrm{s} \) and calculate the distance the log fell in meter. \( \mathrm{g}=9.81 \mathrm{~m} / \mathrm{s}^{2} \) \( v=g \mathbf{t} \) \( d=1 / x g t^{2} \) 8. I drop a quarter from the top of tho towar at the front of Campboll High School and it takes 3.25 seconds to hit the ground, Calculate the velocity in \( \mathrm{m} / \mathrm{s} \) after 1.40 seconds of freefall. How far did the penny fall when ithit the ground? 7. If I drop an peach from the top of one of the tower dorms at UGA, and it takes 7.56 seconds to hit the ground, calculate how tall the building is in meters. 8. You are walking in Paris alongside the Eiffel Tower and suddenly a croissant smacks you on the head and knocks you to the ground. From your handy dandy tourist guidebook you find that the height of the Eiffel Tower is 400.7 m . If you neglect air resistance, calculate how many seconds the croissant dropped before it tagged you on the head. 9. During the latter part of your European vacation, you are hanging out the beach at the gold coast of Spain. As you are laying in your chaise lounge soaking up the warm Mediterranean sun, a large glob of seagull poop hits you in the face. Since you got an "A" in ICPE you are able to estimate the impact velocity at \( 73.7 \mathrm{~m} / \mathrm{s} \). Neglecting air resistance, calculate how high up the seagull was flying when it pooped. 10. If you were to throw a large log over the edge of the Grand Canyon and it took 8.35 seconds to hit the ground, calculate the velocity of the log at impact in \( \mathrm{m} / \mathrm{s} \) and calculate the distance the log fell in meter. Acceleration Calculations

Ask by Weaver Craig. in the United States
Jan 23,2025

Real Tutor Solution

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Answer

- **Log over Grand Canyon (5.65s):** Velocity at impact ≈ 55.43 m/s, distance fallen ≈ 156.58 meters - **Quarter from tower (3.25s):** Velocity after 1.40s ≈ 13.73 m/s, distance fallen ≈ 51.81 meters - **Peach from tower (7.56s):** Building height ≈ 280.34 meters - **Croissant from Eiffel Tower:** Time to fall ≈ 81.69 seconds - **Seagull poop impact:** Height ≈ 276.84 meters - **Log over Grand Canyon (8.35s):** Velocity at impact ≈ 81.91 m/s, distance fallen ≈ 341.99 meters

Solution

Let's solve the problems step by step, using the known conditions and the provided formulas. ### Problem 5: Log over the Grand Canyon 1. **Given:** - Time \( t = 5.65 \) seconds - Acceleration due to gravity \( g = 9.81 \, \mathrm{m/s^2} \) 2. **Calculate the velocity at impact \( v \):** \[ v = g \cdot t \] 3. **Calculate the distance fallen \( d \):** \[ d = \frac{1}{2} g t^2 \] Now, let's perform the calculations for both \( v \) and \( d \). ### Problem 8: Quarter dropped from a tower 1. **Given:** - Time \( t = 3.25 \) seconds - Acceleration due to gravity \( g = 9.81 \, \mathrm{m/s^2} \) 2. **Calculate the velocity after \( 1.40 \) seconds of free fall:** \[ v = g \cdot t \] where \( t = 1.40 \) seconds. 3. **Calculate the distance fallen when it hit the ground:** \[ d = \frac{1}{2} g t^2 \] ### Problem 7: Peach dropped from a tower 1. **Given:** - Time \( t = 7.56 \) seconds - Acceleration due to gravity \( g = 9.81 \, \mathrm{m/s^2} \) 2. **Calculate the height of the building \( d \):** \[ d = \frac{1}{2} g t^2 \] ### Problem 8: Croissant dropped from the Eiffel Tower 1. **Given:** - Height \( d = 400.7 \) m - Acceleration due to gravity \( g = 9.81 \, \mathrm{m/s^2} \) 2. **Calculate the time \( t \) it took to fall:** \[ d = \frac{1}{2} g t^2 \] ### Problem 9: Seagull poop impact velocity 1. **Given:** - Impact velocity \( v = 73.7 \, \mathrm{m/s} \) - Acceleration due to gravity \( g = 9.81 \, \mathrm{m/s^2} \) 2. **Calculate the height \( d \) from which the seagull dropped:** \[ d = \frac{v^2}{2g} \] ### Problem 10: Log over the Grand Canyon (second instance) 1. **Given:** - Time \( t = 8.35 \) seconds - Acceleration due to gravity \( g = 9.81 \, \mathrm{m/s^2} \) 2. **Calculate the velocity at impact \( v \):** \[ v = g \cdot t \] 3. **Calculate the distance fallen \( d \):** \[ d = \frac{1}{2} g t^2 \] Now, let's perform all the calculations in parallel. Calculate the value by following steps: - step0: Calculate: \(9.81\times 1.4\) - step1: Multiply the numbers: \(13.734\) Calculate or simplify the expression \( 9.81 * 8.35 \). Calculate the value by following steps: - step0: Calculate: \(9.81\times 8.35\) - step1: Multiply the numbers: \(81.9135\) Calculate or simplify the expression \( 9.81 * 5.65 \). Calculate the value by following steps: - step0: Calculate: \(9.81\times 5.65\) - step1: Multiply the numbers: \(55.4265\) Calculate or simplify the expression \( (400.7 * 2) / 9.81 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(400.7\times 2\right)}{9.81}\) - step1: Remove the parentheses: \(\frac{400.7\times 2}{9.81}\) - step2: Multiply the numbers: \(\frac{801.4}{9.81}\) - step3: Convert the expressions: \(\frac{\frac{4007}{5}}{\frac{981}{100}}\) - step4: Multiply by the reciprocal: \(\frac{4007}{5}\times \frac{100}{981}\) - step5: Reduce the numbers: \(4007\times \frac{20}{981}\) - step6: Multiply: \(\frac{4007\times 20}{981}\) - step7: Multiply: \(\frac{80140}{981}\) Calculate or simplify the expression \( 0.5 * 9.81 * (8.35^2) \). Calculate the value by following steps: - step0: Calculate: \(0.5\times 9.81\times 8.35^{2}\) - step1: Convert the expressions: \(0.5\times 9.81\left(\frac{167}{20}\right)^{2}\) - step2: Multiply the terms: \(4.905\left(\frac{167}{20}\right)^{2}\) - step3: Convert the expressions: \(\frac{981}{200}\left(\frac{167}{20}\right)^{2}\) - step4: Multiply the numbers: \(\frac{27359109}{80000}\) Calculate or simplify the expression \( 0.5 * 9.81 * (7.56^2) \). Calculate the value by following steps: - step0: Calculate: \(0.5\times 9.81\times 7.56^{2}\) - step1: Convert the expressions: \(0.5\times 9.81\left(\frac{189}{25}\right)^{2}\) - step2: Multiply the terms: \(4.905\left(\frac{189}{25}\right)^{2}\) - step3: Convert the expressions: \(\frac{981}{200}\left(\frac{189}{25}\right)^{2}\) - step4: Multiply the numbers: \(\frac{35042301}{125000}\) Calculate or simplify the expression \( 0.5 * 9.81 * (5.65^2) \). Calculate the value by following steps: - step0: Calculate: \(0.5\times 9.81\times 5.65^{2}\) - step1: Convert the expressions: \(0.5\times 9.81\left(\frac{113}{20}\right)^{2}\) - step2: Multiply the terms: \(4.905\left(\frac{113}{20}\right)^{2}\) - step3: Convert the expressions: \(\frac{981}{200}\left(\frac{113}{20}\right)^{2}\) - step4: Multiply the numbers: \(\frac{12526389}{80000}\) Calculate or simplify the expression \( 0.5 * 9.81 * (3.25^2) \). Calculate the value by following steps: - step0: Calculate: \(0.5\times 9.81\times 3.25^{2}\) - step1: Convert the expressions: \(0.5\times 9.81\left(\frac{13}{4}\right)^{2}\) - step2: Multiply the terms: \(4.905\left(\frac{13}{4}\right)^{2}\) - step3: Convert the expressions: \(\frac{981}{200}\left(\frac{13}{4}\right)^{2}\) - step4: Multiply the numbers: \(\frac{165789}{3200}\) Calculate or simplify the expression \( (73.7^2) / (2 * 9.81) \). Calculate the value by following steps: - step0: Calculate: \(\frac{73.7^{2}}{\left(2\times 9.81\right)}\) - step1: Remove the parentheses: \(\frac{73.7^{2}}{2\times 9.81}\) - step2: Convert the expressions: \(\frac{\left(\frac{737}{10}\right)^{2}}{2\times 9.81}\) - step3: Multiply the numbers: \(\frac{\left(\frac{737}{10}\right)^{2}}{19.62}\) - step4: Convert the expressions: \(\frac{\left(\frac{737}{10}\right)^{2}}{\frac{981}{50}}\) - step5: Evaluate the power: \(\frac{\frac{737^{2}}{100}}{\frac{981}{50}}\) - step6: Multiply by the reciprocal: \(\frac{737^{2}}{100}\times \frac{50}{981}\) - step7: Reduce the numbers: \(\frac{737^{2}}{2}\times \frac{1}{981}\) - step8: Multiply the fractions: \(\frac{737^{2}}{2\times 981}\) - step9: Multiply: \(\frac{737^{2}}{1962}\) Here are the detailed calculations and results for each problem: ### Problem 5: Log over the Grand Canyon 1. **Velocity at impact \( v \):** \[ v = 9.81 \cdot 5.65 \approx 55.43 \, \mathrm{m/s} \] 2. **Distance fallen \( d \):** \[ d = \frac{1}{2} \cdot 9.81 \cdot (5.65^2) \approx 156.58 \, \mathrm{m} \] ### Problem 8: Quarter dropped from a tower 1. **Velocity after \( 1.40 \) seconds:** \[ v = 9.81 \cdot 1.40 \approx 13.73 \, \mathrm{m/s} \] 2. **Distance fallen when it hit the ground:** \[ d = \frac{1}{2} \cdot 9.81 \cdot (3.25^2) \approx 51.81 \, \mathrm{m} \] ### Problem 7: Peach dropped from a tower 1. **Height of the building:** \[ d = \frac{1}{2} \cdot 9.81 \cdot (7.56^2) \approx 280.34 \, \mathrm{m} \] ### Problem 8: Croissant dropped from the Eiffel Tower 1. **Time it took to fall:** \[ t = \sqrt{\frac{2 \cdot 400.7}{9.81}} \approx 81.69 \, \mathrm{s} \] ### Problem 9: Seagull poop impact velocity 1. **Height from which the seagull dropped:** \[ d = \frac{(73.7)^2}{2 \cdot 9.81} \approx 276.84 \, \mathrm{m} \] ### Problem 10: Log over the Grand Canyon (second instance) 1. **Velocity at impact \( v \):** \[ v = 9.81 \cdot 8.35 \approx 81.91 \, \mathrm{m/s} \] 2. **Distance fallen \( d \):** \[ d = \frac{1}{2} \cdot 9.81 \cdot (8.35^2) \approx 341.99 \, \mathrm{m} \] ### Summary of Results: - **Log over Grand Canyon (5.65s):** \( v \approx 55.43 \, \mathrm{m/s}, d \approx 156.58 \, \mathrm{m} \) - **Quarter from tower (3.25s):** \( v \approx 13.73 \, \mathrm{m/s}, d \approx 51.81 \, \mathrm{m} \) - **Peach from tower (7.56s):** \( d \approx 280.34 \, \mathrm{m} \) - **Croissant from Eiffel Tower:** \( t \approx 81.69 \, \mathrm{s} \) - **Seagull poop impact:** \( d \approx 276.84 \, \mathrm{m} \) - **Log over Grand Canyon (8.35s):** \( v \approx 81.91 \, \mathrm{m/s}, d \approx 341.99 \, \mathrm{m} \)

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Did you know that the Grand Canyon has a depth of over a mile? When dropping that log, the laws of physics make it a thrilling experiment! For example, if it takes 8.35 seconds to hit the ground, using the formula \( v = g \cdot t \), we find that the log's impact velocity is about \( 81.92 \, \mathrm{m/s} \) (since \( g = 9.81 \, \mathrm{m/s}^2 \)). To find the distance fallen, you would calculate \( d = \frac{1}{2} g t^2 \), giving a depth of approximately \( 343.25 \, \mathrm{m} \). If you have a soft spot for the classic physics calculations, you might want to explore Galileo's experiments with falling objects! He famously dropped spheres from the Leaning Tower of Pisa and disproved the ancient belief that heavier objects fall faster. Such experiments laid the groundwork for our understanding of gravitational acceleration, making physics not just a subject, but a fascinating journey through history!

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