If the ball could continue to travel below it’s original level (through an appropriately shaped hole in the ground), then negative values of y corresponding to times greater than 6.04s would be possible. Compare the balls position and velocity 8.00s after the start of its flight. (Answers c =178m. Y= 76.8m, Vx= 22.2m/s, Vy= -48.8m/sBoth images have background information to help solve the practice problem. (The correct answer is also provided) just need help with working it out. the air. Ccapturing up to 250,000 frames per second. These data reveal that the spores are ejected with accelerations ofup to 1,800,000 m/s, or about 180,000g. These are the largest natural accelerations ever measured.EXAMPLE 3.4 A home-run hitNow let's consider a projectile problem in which the initial velocity is specified in terms of a magnitudeand an angle. Suppose a home-run baseball is hit with an initial speed vo= 37.0 m/s at an initial angle80 = 53.1°. (a) Find the ball's position, and the magnitude and direction of its velocity, when t = 2.00 s.(b) Find the time the ball reaches the highest point of its flight, and find its height h at that point. (c) Find thehorizontal range R (the horizontal distance from the starting point to the point where the ball hits the ground).SOLUTIONSET UP The ball is probably struck a meter or so above ground level,but we'll ignore this small distance and assume that it starts at groundlevel (yo = 0). Figure 3.15 shows our sketch. We place the origin ofcoordinates at the starting point, so xo = 0. The components of the initialvelocity areVox= vo cos 0o = (37.0 m/s) (0.600) = 22.2 m/s,Voy vo sin 0o = (37.0 m/s) (0.800) = 29.6 m/s.y (m)VoyVo = 37.0 m/s000¹ VoxT₁Vy=0h=?di TOROA FIGURE 3.15 Our sketch for this problem.signa→→→→x (m)x = R = ?t₂ = ?SOLVE Part (a): We ignore the effects of the air. We want to find x, y,Ux, and u, at time t = 2.00 s. We substitute these values into Equations 3.8through 3.11, along with the value t = 2.00 s. The coordinates att = 2.00 s arex = Voxt = (22.2 m/s) (2.00 s) = 44.4 m,y = Voyt - - 81²(29.6 m/s) (2.00 s) (9.80 m/s²) (2.00 s)² = 39.6 m.The components U, and u, of the velocity vector at time t = 2.00 s areUx= Vox = 22.2 m/s,BurUy = Voy gt = 29.6 m/s + (-9.80 m/s²) (2.00 s)= 10.0 m/s.Video Tutor SolutionpidenaltThe magnitude and direction of the velocity vector at time t = 2.00 sare, respectively,suim ontledV = V√(22.2 m/s)2 + (10.0 m/s)² = 24.3 m/s,0 = tan+ 0,²10.0 m/s22.2 m/s= tan ¹0.450 = 24.2°.CONTINUED 76 CHAPTER 3 Motion in a PlanePart (b): At the highest point in the ball's path, the vertical velocitycomponent u, is zero. When does this happen? Call that time ₁; then,using uy = Voy gt, we find thatUy = 0 = 29.6 m/s - (9.80 m/s²)₁,t₁ = 3.02 s.The height h at this time is the value of y when t = 3.02 s. We useEquation 3.11,y = yo voyt + (-8)1²,to obtainh = 0 + (29.6 m/s) (3.02 s) + (-9.80 m/s²) (3.02 s)²= 44.7 m.Alternatively, we can use the constant-acceleration formula,v² = voy² + 2a(y-yo) = voy² + 2(-8)(y - yo).At the highest point, uy = 0 and y = h. Substituting these values in,along with yo = 0, we find that0 = (29.6 m/s)22(9.80 m/s²)h,h = 44.7 m.This is roughly half the height of the Statue of Liberty above the play-ing field.Part (c): To find the range R, we start by asking when the ball hits theground. This occurs when y = 0. Call that time 12; then, from Equa-tion 3.11,y = 0 = (29.6 m/s)t2 + (-9.80 m/s²)t₂².This is a quadratic equation for 12; it has two roots:12 = 0 and 1₂ = 6.04 s.EXAMPLE 3.5 Range andThere are two times at which y = 0: 12 = 0 is the time the ball leavesthe ground, and 12 = 6.04 s is the time of its return. The latter is exactlytwice the time the ball takes to reach the highest point, so the time ofdescent equals the time of ascent. (This is always true if the starting andending points are at the same elevation and air resistance is ignored.We'll prove it in Example 3.5.)The range R is the value of x when the ball returns to the ground-that is, when t = 6.04 s:R = Vox 12 = (22.2 m/s) (6.04 s) = 134 m.For comparison, the distance from home plate to center field at Pittsburgh'sPNC Park is 399 ft (about 122 m). If we could ignore air resistance, the ballreally would be a home run.LumaldAt the instant the ball returns to y = 0, its vertical component ofvelocity isUy29.6 m/s + (-9.80 m/s²) (6.04 s) = -29.6 m/s.That is, uy has the same magnitude as the initial vertical velocity voy butthe opposite sign. Since u, is constant, the angle = -53.1° (below thehorizontal) at this point is the negative of the initial angle 00 = 53.1°.REFLECT The actual values of the maximum height h and the range Rare substantially lower than the values we've found because air resis-tance is not negligible. In fact, the range of a batted ball is substantiallygreater (on the order of 10 m for a home-run ball) in Denver than inPittsburgh because the density of air is almost 20% lower in Denver.Practice Problem: If the ball could continue to travel below its origi-nal level (through an appropriately shaped hole in the ground), thennegative values of y corresponding to times greater than 6.04 s wouldbe possible. Compute the ball's position and velocity 8.00 s after thestart of its flight. Answers: x= 178 m, y = -76.8 m, v, = 22.2 m/s,Uy -48.8 m/s.

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Description: If the ball could continue to travel below it’s original level (through an appropriately shaped hole in the ground), then negative values of y corresponding to times greater than 6.04s would be possible. Compare the balls position and velocity 8.00s

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