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The coin is released 47.0 m above the ground. How much time is required for the coin to fall for the ground? What is it’s velocity just before it strikes the ground? Answer: 3.10s , -30.4 m/s The images have background info to help solve the practice problem.

The coin is released 47.0 m above the ground. How much time is required for the coin to fall for the ground? What is it’s velocity just before it strikes the ground? Answer: 3.10s , -30.4 m/s The images have background info to help solve the practice problem.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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The coin is released 47.0 m above the ground. How much time is required for the coin to fall for the ground? What is it’s velocity just before it strikes the ground? Answer: 3.10s , -30.4 m/s The images have background info to help solve the practice problem.
67 Typicanju y axis points upward. For such a cool -g. The im ject has only a y component, which is given by ay = that the acceleration vector points downward. We know this should be true because throw an object straight up, it will start with a positive initial velocity that will ever become negative as the object reaches its maximum height and then begins to fa down. So the acceleration transforms a positive velocity into a negative one-he- acceleration itself must be negative. Since free fall is constant accelerated motion, in fact, use Equations 2.6, 2.10, and 2.11 to describe the motion with only min fications. First, we need to change all of the x's in those equations to y's becau assuming that the free-fall motion is along the y direction. Second, we need to with ay = -g. This results in the following free-fall equations: Equations of motion for free fall Uy = Voy - gt y = yo + voyt - 1/181² v² = voy 2g(y - yo) Notes: ● g = +9.80 m/s² in these equations. ● yo and voy are the position and the velocity of the object at t = 0. EXAMPLE 2.10 Falling euro in Pisa Our first example of free-fall motion is very simple. A 2-euro coin is dropped from the Leaning Tower of Pisa. It starts from rest and falls freely. Ignore the effects of air resistance, and compute its position and velocity after 1.00 s, 2.00 s, and 3.00 s.
Transcribed Image Text:67 Typicanju y axis points upward. For such a cool -g. The im ject has only a y component, which is given by ay = that the acceleration vector points downward. We know this should be true because throw an object straight up, it will start with a positive initial velocity that will ever become negative as the object reaches its maximum height and then begins to fa down. So the acceleration transforms a positive velocity into a negative one-he- acceleration itself must be negative. Since free fall is constant accelerated motion, in fact, use Equations 2.6, 2.10, and 2.11 to describe the motion with only min fications. First, we need to change all of the x's in those equations to y's becau assuming that the free-fall motion is along the y direction. Second, we need to with ay = -g. This results in the following free-fall equations: Equations of motion for free fall Uy = Voy - gt y = yo + voyt - 1/181² v² = voy 2g(y - yo) Notes: ● g = +9.80 m/s² in these equations. ● yo and voy are the position and the velocity of the object at t = 0. EXAMPLE 2.10 Falling euro in Pisa Our first example of free-fall motion is very simple. A 2-euro coin is dropped from the Leaning Tower of Pisa. It starts from rest and falls freely. Ignore the effects of air resistance, and compute its position and velocity after 1.00 s, 2.00 s, and 3.00 s.
of an object as held the Oneously) alileo ar- ent of its ented by r resis- -articu- shows fall at of the tant. nce, call to or 2 e SOLUTION SET UP Figure 2.25 shows the diagram we sketch. For a coordinate system, we choose a vertical axis with the upward direction to be posi- tive, and we place the origin at the starting point. The position, velocity, and acceleration have only y components. The initial value yo of the co- ordinate y and the initial y component of velocity voy are both zero. The acceleration is downward (in the negative y direction), so the constant y component of acceleration is a, = -g = -9.80 m/s². (Remember that, by definition, g itself is always positive.) SOLVE From Equations 2.13 and 2.12 we have y = vot + a,t² = 0+ (-8)² = (-4.90 m/s²) 1², Uy = Voy at = 0 + (-g)t = (-9.80 m/s²)t. Using subscript 1 for the first time, we find that when t = t₁ = 1.00 s, y1 = (-4.90 m/s2) (1.00 s)² = -4.90 m and v₁y = (-9.80 m/s²) (1.00 s) = -9.80 m/s. The coin is therefore 4.90 m be- low the origin (because y is negative) and has a downward velocity (vy is negative) with magnitude 9.80 m/s. The positions and velocities at 2.00 s and 3.00 s are found in the same way. The results are -19.6 m and -19.6 m/s at t = 2.00 s and -44.1 m and -29.4 m/s at t = 3.00 s. REFLECT All our values of y and v, are negative because we have cho- sen the positive y axis to be upward. We could just as well have chosen the downward direction to be positive. In that case, the y component of acceleration would have been ay = g = +9.80 m/s². The crucial point is that you must decide at the start which direction is positive, and then stick with your decision. The Leaning Tower zo sift CAIRO! A FIGURE 2.25 2.6 Freely Falling Objects Our sketch for the problem Y -to= 0, yo=0 Yo = 0 M₁y =? T1=1.00 s, y₁ = ? 9--9--9.80 m/s² +₂=2.00 s, Y₂ = ? 13=3.00 S, Y3= ? 49 Luby? | Practice Problem: The coin is released 47.0 m above the ground. How much time is required for the coin to fall to the ground? What is its velocity just before it strikes the ground? Answers: 3.10 s, -30.4 m/s.
Transcribed Image Text:of an object as held the Oneously) alileo ar- ent of its ented by r resis- -articu- shows fall at of the tant. nce, call to or 2 e SOLUTION SET UP Figure 2.25 shows the diagram we sketch. For a coordinate system, we choose a vertical axis with the upward direction to be posi- tive, and we place the origin at the starting point. The position, velocity, and acceleration have only y components. The initial value yo of the co- ordinate y and the initial y component of velocity voy are both zero. The acceleration is downward (in the negative y direction), so the constant y component of acceleration is a, = -g = -9.80 m/s². (Remember that, by definition, g itself is always positive.) SOLVE From Equations 2.13 and 2.12 we have y = vot + a,t² = 0+ (-8)² = (-4.90 m/s²) 1², Uy = Voy at = 0 + (-g)t = (-9.80 m/s²)t. Using subscript 1 for the first time, we find that when t = t₁ = 1.00 s, y1 = (-4.90 m/s2) (1.00 s)² = -4.90 m and v₁y = (-9.80 m/s²) (1.00 s) = -9.80 m/s. The coin is therefore 4.90 m be- low the origin (because y is negative) and has a downward velocity (vy is negative) with magnitude 9.80 m/s. The positions and velocities at 2.00 s and 3.00 s are found in the same way. The results are -19.6 m and -19.6 m/s at t = 2.00 s and -44.1 m and -29.4 m/s at t = 3.00 s. REFLECT All our values of y and v, are negative because we have cho- sen the positive y axis to be upward. We could just as well have chosen the downward direction to be positive. In that case, the y component of acceleration would have been ay = g = +9.80 m/s². The crucial point is that you must decide at the start which direction is positive, and then stick with your decision. The Leaning Tower zo sift CAIRO! A FIGURE 2.25 2.6 Freely Falling Objects Our sketch for the problem Y -to= 0, yo=0 Yo = 0 M₁y =? T1=1.00 s, y₁ = ? 9--9--9.80 m/s² +₂=2.00 s, Y₂ = ? 13=3.00 S, Y3= ? 49 Luby? | Practice Problem: The coin is released 47.0 m above the ground. How much time is required for the coin to fall to the ground? What is its velocity just before it strikes the ground? Answers: 3.10 s, -30.4 m/s.
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