I just need the practice problem solved with work shown. The rest of the image provides the information needed to solve the problem. The correct answer is also provided to check work. (The practice problem is on the second image at the bottom) BodWEXAMPLE 2.11 A ball on the roofNow let's examine the more complex case of a ball being thrown vertically upward. Even though the ballis initially moving upward, it is, nevertheless, in free fall and we can use Equations 2.12, 2.13, and 2.14to analyze its motion. Suppose you throw a ball vertically upward from the flat roof of a tall building. Theball leaves your hand at a point even with the roof railing, with an upward velocity of 15.0 m/s. On its wayback down, it just misses the railing. Find (a) the position and velocity of the ball 1.00 s and 4.00 s after itleaves your hand; (b) the velocity of the ball when it is 5.00 m above the railing; and (c) the maximum heightreached and the time at which it is reached. Ignore the effects of the air. 50CHAPTER 2 Motion Along a Straight LineSOLUTIONSET UP As shown in Figure 2.27, we place the origin at the level ofthe roof railing, where the ball leaves your hand, and we take the posi-tive direction to be upward. Here's what we know: The initial positionYo is zero, the initial velocity voy (y component) is +15.0 m/s, and theacceleration (y component) is a, = -9.80 m/s².The ball actually moves straight upthen straight down; we showa U-shaped path for clarity.,1 = 0, VoyA FIGURE 2.27५<= ?1 = 1.00 s५ = ?7 = ?==15.0 m/sandy = 01 = ?U = ?t =4.00 sU₂ = ?T3 =The velocity uy at any position y is given by2v² = voy² + 2a,(y - Yo)y = 5.00 my = 0ay = -8= -9.80 m/s²y = ?DanhipseSOLVE What equations do we have to work with? The velocity v, atany time t isVy = Voy + ayt = 15.0 m/s + (-9.80 m/s²)t.The position y at any time t isy = voyt + ¼a₂t² = (15.0 m/s)t + (-9.80 m/s²)t².(15.0 m/s)2 + 2(-9.80 m/s²) (y - 0).Part (a): When t = 1.00 s, the first two equations give y=Vy+5.20 m/s. The ball is 10.1 m above the origin (y is positive),and it has an upward velocity (v is positive) of 5.20 m/s (less than theinitial velocity of 15.0 m/s, as expected). When t = 4.00 s, the sameequations give= +10.1m,y = -18.4 m, uy = -24.2 m/s.Thus, at time t = 4.00 s, the ball has passed its highest point and is3.4 m below the origin (y is negative). It has a downward velocity (v, isgative) with magnitude 24.2 m/s (greater than the initial velocity, ase should expect). Note that, to get these results, we don't need to findhighest point the ball reaches or the time at which it is reached. Thenations of motion give the position and velocity at any time, whetherball is on the way up or on the way down.=+5.00Part (b): When the ball is 5.00 m above the origin, ywe use our third equation to find the velocity u, at this point:127 m²/s²,v² = (15.0 m/s)² + 2(-9.80 m/s²) (5.00 m)U₂ = ± 11.3 m/s.passes the point yWe get two values of Uy, one positive and one negative. That is, the ball+5.00 m twice, once on the way upand again onthe way down. The velocity on the way up is +11.3 m/s, and on theway down it is -11.3 m/s.Part (c): At the highest point, the ball's velocity is momentarily zero(vy=0); it has been going up (positive v,) and is about to start goingdown (negative v.). From our third equation, we have0 = (15.0 m/s)² - (19.6 m/s²)y,111.5 m. We can nowand the maximum height (where uy = 0) is y =find the time when the ball reaches its highest point from Equation 2.12,setting v, = 0:y (m)Alternative Solution: Alternatively, to find the maximum height, wemay ask first when the maximum height is reached. That is, at what= 0 when t = 1.53 s. Substi-= 0? As we just found, vyvalue of t is vytuting this value of t back into the equation for y, we find thaty = (15 m/s)(1.53 s) + (−9.8 m/s²) (1.53 s)² = 11.5 m.15=REFLECT Although at the highest point the velocity is momentarilyzero, the acceleration at that point is still -9.80 m/s². The ball stopsfor an instant, but its velocity is continuously changing, from positivevalues through zero to negative values. The acceleration is constantthroughout; it is not zero when uy = 0!Figure 2.28 shows graphs of position and velocity as functions oftime for this problem. Note that the graph of uy versus t has a constantnegative slope. Thus, the acceleration is negative on the way up, at thehighest point, and on the way down.93103Practice Problem: If the building is 30.0 m high, how long does it takefor the ball to hit the ground? What velocity is the ball traveling at whenit hits the ground? Answers: 4.44 s, -28.5 m/s.010sday 5 RUE-5-100 = 15.0 m/s + (-9.80 m/s²)t,t = 1.53 s.-15 -=-201 234t(s)lladUy (m/s)1510•50↑ ÷m. Now-5---1The acceleration isconstant (straight line)and negative (negativeslope).2iisd1-10-15-20-25(a)(b)▲ FIGURE 2.28 (a) Position and (b) velocity as functions of time foball thrown upward with an initial velocity of 15 m/s.3 4t(s)

Answered: EX No is in to a ball back leave reach

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)...