1) A physics book slides off a horizontal tabletop with a speed of 3.80 . It str
ID: 2237924 • Letter: 1
Question
1) A physics book slides off a horizontal tabletop with a speed of 3.80 . It strikes the floor in 0.350 . Ignore air resistance. Find the horizontal component of the book's velocity, just before the book reaches the floor. Find the magnitude of book's velocity, just before the book reaches the floor. Find the direction of book's velocity, just before the book reaches the floor. 2) A major leaguer hits a baseball so that it leaves the bat at a speed of 28.5 and at an angle of 36.5 above the horizontal. You can ignore air resistance. At what two times is the baseball at a height of 8.50 m above the point at which it left the bat? Calculate the horizontal component of the baseball's velocity at each of the two times you found in part A. Calculate the vertical component of the baseball's velocity at each of the two times you found in part A. What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat? What is the direction of the baseball's velocity when it returns to the level at which it left the bat? 3) A baseball thrown at an angle of 65.0 degrees above the horizontal strikes a building 18.0 m away at a point 7.00 m above the point from which it is thrown. Ignore air resistance. Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown). Find the magnitude of the velocity of the baseball just before it strikes the building. Find the direction of the velocity of the baseball just before it strikes the building. 4) A curving freeway exit has a radius of 48.5 and a posted speed limit of 35 . What is your radial acceleration (in m/s^2 ) if you take this exit at the posted speed? What is your radial acceleration (in m/s^2 ) if you take the exit at a speed of 50 ?Explanation / Answer
1.height = 0.6m horizontal distcance = 0.385m hor component = 1.10m/s vert component = -3.43m/s 3.6m/s 72.2deg below horizontal 2.Vertical component of its velocity, v = (28.5) sin 36.9 degrees = 17.11 m/s For vertical motion, y = vt - 4.9t^2 9.5 = 17.11t - 4.9t^2 4.9t^2 - 17.11t + 9.5 = 0 t = 0.69 s or 2.9 s 3.(a). x = Vo t cos ? t = x/(Vo cos ?) y = Vo t sin ? +
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