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Chapter 2 Examples

Chapter 2 Examples. Problem 2.46. An egg is thrown vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point 50 m below its starting point 5.0 s after it leaves the thrower’s hand. Ignore air resistance.

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Chapter 2 Examples

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  1. Chapter 2 Examples

  2. Problem 2.46 An egg is thrown vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point 50 m below its starting point 5.0 s after it leaves the thrower’s hand. Ignore air resistance. • What is the initial speed of the egg? • How high does it rise above its starting point? • What is the magnitude of its velocity at its highest point? • What are the magnitude and direction of its acceleration at the highest point?

  3. Step 1: Draw it h, height above building 50 m in 5 s

  4. Make some definitions • Let t=5 s • Let vo= initial velocity, in positive direction • Let g=9.8 m/s2 , acceleration due to gravity, in negative direction. • total distance =-50 m

  5. Apply Definitions and Solve • Let t=5 s • Let vo= initial velocity, in positive direction • Let g=9.8 m/s2 , acceleration due to gravity, in negative direction. This will be “a” • total distance =-50 m =y(t)-yo

  6. Part B) Find height above bldg • v0=14.5 m/s • At top of the arc, vf=0 • At t=0, let y0=0

  7. Part B) Find height above bldg • v0=14.5 m/s • At top of the arc, vf=0 • At t=0, let y0=0 • Let yf=h

  8. Parts C) and D) • At the top of the arc, v=0 m/s • Always in this problem a=-g=-9.8 m/s2

  9. Problem 2.61 A car 3.5 m in length and traveling at a constant speed of 20 m/s is approaching an intersection which is 20 m wide. The light turns yellow when the front of the car is 50 m from the beginning of the intersection. If the driver steps on the brake, the car will slow at -3.8 m/s2. If the driver steps on the gas pedal, the car will accelerate at 2.3 m/s2. The light will be yellow for 3 seconds. Ignore reaction time. To avoid being in the intersection when the light changes to red, accelerate or brake?

  10. Step 1: Draw It! 3.5 m In order for the car to run the green light, It must traverse 3.5+50+20 m=73.5 m so that no part of the car is in the intersection Braking is easier, it must traverse less than 50 m 50 m 20 m

  11. Our Options • We know that • t=3 s • a is either 2.3 m/s2 or -3.8 m/s2 • vo=20 m/s • Total distance is either 73.5 m or 50 m

  12. Our Options • We know that • t=3 s • a is either 2.3 m/s2 or -3.8 m/s2 • vo=20 m/s • Total distance is either >73.5 m or <50 m Braking Run It !

  13. Problem 2.77 A physics student with too much free time drops a water melon from the roof of a building. He hears the sound of the watermelon going “splat” 2.5 s later. How high is the building? The speed of sound is 340 m/s and ignore air resistance.

  14. Step 1: Draw It! h Splat!

  15. Some Hard Thinkin’ • The melon experiences an acceleration due to gravity. The student merely dropped it, so its initial velocity was 0. • The sound wave is unaffected by gravity so it moves with constant velocity from the ground toward the student. • These are two separate events with a total time of 2.5 s

  16. Some Hard Thinkin’ part 2 • The equation for the distance that the melon traverses is y=-1/2*g*(t1)2 where y= height of bldg and t1 is the time for the fall. • The equation of distance for the sound wave is y=vs*t2 where vs = speed of sound =340 m/s • The total time for all this to transpire is 2.5 s or 2.5 s =t1+t2

  17. Solving

  18. Problem 2.89 A painter is standing on scaffolding that is raised at a constant speed. As he travels upward, he accidentally nudges a paint can off the scaffolding and it falls 15 m to the ground. You are watching and measure with your stopwatch that it takes 3.25 s for the can to reach the ground. • What is the speed of the can just before it hits the ground? • Another painter is standing on a ledge with his hands 4 m above the can when it falls. He has lightning-fast reflexes and can catch if at all possible. Does he get a chance?

  19. Part a) Make some definitions • Let t=3.25 s • Let vo= initial velocity, in positive direction • Let g=9.8 m/s2 , acceleration due to gravity, in negative direction. • total distance =-15 m

  20. Make some definitions • We must solve this equation for vo • And then we must use this equation to solve for the velocity

  21. Solving

  22. Part B) Make some definitions • The word “falls” is slightly misleading, the can first rises in the air and then falls to the ground. • What it is asking for is how high the can flies above the release point; it must be greater than 4 m • So we are solving for a total “y” displacement (yf-yo) • At the top of the arc, vf=0 • Let vo= initial velocity, in positive direction, 11.31 m/s • Let g=9.8 m/s2 , acceleration due to gravity, in negative direction.

  23. Part B) Make some definitions • What it is asking for is how high the can flies above the release point; it must be greater than 4 m • So we are solving for a total “y” displacement (yf-yo) • At the top of the arc, vf=0 • Let vo= initial velocity, in positive direction, 11.31 m/s • Let g=9.8 m/s2 , acceleration due to gravity, in negative direction. Yes, he can catch it

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