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/sThe images have background info to help solve the practice problem. 67Typicanjuy axis points upward. For such a cool-g. The imject has only a y component, which is given by ay =that the acceleration vector points downward. We know this should be true becausethrow an object straight up, it will start with a positive initial velocity that will everbecome negative as the object reaches its maximum height and then begins to fadown. 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 minfications. First, we need to change all of the x's in those equations to y's becauassuming that the free-fall motion is along the y direction. Second, we need towith ay = -g. This results in the following free-fall equations:Equations of motion for free fallUy = Voy - gty = yo + voyt - 1/181²v² = voy2g(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 PisaOur first example of free-fall motion is very simple. A 2-euro coin is dropped from the Leaning Tower ofPisa. It starts from rest and falls freely. Ignore the effects of air resistance, and compute its position andvelocity after 1.00 s, 2.00 s, and 3.00 s. of an objectas held theOneously)alileo ar-ent of itsented byr resis--articu-showsfall atof thetant.nce,calltoor2eSOLUTIONSET UP Figure 2.25 shows the diagram we sketch. For a coordinatesystem, 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. Theacceleration is downward (in the negative y direction), so the constanty component of acceleration is a, = -g = -9.80 m/s². (Rememberthat, by definition, g itself is always positive.)SOLVE From Equations 2.13 and 2.12 we havey = 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 whent = 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 (vyis negative) with magnitude 9.80 m/s.The positions and velocities at 2.00 s and 3.00 s are found in thesame 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 chosenthe downward direction to be positive. In that case, the y component ofacceleration would have been ay = g = +9.80 m/s². The crucial pointis that you must decide at the start which direction is positive, and thenstick with your decision.The Leaning Towerzo siftCAIRO!A FIGURE 2.252.6 Freely Falling ObjectsOur sketch for the problemY-to= 0, yo=0Yo = 0M₁y =? T1=1.00 s, y₁ = ?9--9--9.80 m/s²+₂=2.00 s, Y₂ = ?13=3.00 S, Y3= ?49Luby? |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? Whatis its velocity just before it strikes the ground? Answers: 3.10 s,-30.4 m/s.

Question

Accepted Answer

Answered: Image /qna-images/answer/3e6c0d20-cd15-44ff-aa74-cc21113b42e5.jpg

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.