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What is the car's vertical velocity

  1. Sep 12, 2005 #1
    I am having problems with the following questions. Any help would be greatly appreciated!
    19) Automobile travels at constant speed of 60 km/hr (16.7m/s), 800 m along a straight highway that is inclined 5 degrees to the horizontal. An observer notes only the vertical motion of hte car. What is the car's vertical velocity and vertical travel distance.
    Part A: vy= vsin5= 16.7 x sin5 = 1.46 m/s
    Part B: I dont know which equation to use because the problem doesn't state a given amount of time. I dont know how to figure out the time from this but I'm assuming once I get that number I can put it into the equation
    y= (vsin5)t
    80)A pitcher throws a fastball horizontally at a speed of 140 km/hr (38.9m/s) toward home plate, 18.4 m away. a) If the batter's combined reaction and swing times total .350sec, how long can the batter watch the ball after it has left the pitchers hand before swinging? b) In traveling to the plate, how far does the ball drop from its original horizontal line?
    Part A: I calculated 38.9 m/s divided by 18.4 m to get 2.11 sec or the time it takes for the ball to reach home plate from the pitcher. I then subtracted 2.11 sec- .350 sec = 1.76 sec for the time the batter can watch.
    Part B: I tried substituting my numbers into the equation x=vxt = 38.9 m/s x 1.76 sec = 68.5 m but I dont think that this number looks accurate.
  2. jcsd
  3. Sep 12, 2005 #2


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    19 B: How long does it take the car to go 800 m at 16.7 m/s? That's the time the car is going up the slope. However, it is easier to apply the same trigonometry to the distance: the slope is 5 degrees and the hypotenuse is 800 m.

    80 A:?? 38.9 m/s divided by 18.4 m is 2.11 "1/sec". Check your units.
    B: No, the ball does not dip 38.9 m! The pitcher did NOT throw the ball straight down! You are trying to find the vertical distance and the initial speed you were given is horizontal. Take the acceleration due to gravity into account.
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