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. 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?

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

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

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

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

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

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

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?

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

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

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 !

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.

Step 1: Draw It! h Splat!

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

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

Solving

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?

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

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

Solving

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.

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