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Homework Help: Speed of a ball

  1. Dec 14, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown straight upward from the ground with a speed of 30 m/s. What is the average speed of the ball in the first 5 seconds?


    2. Relevant equations
    VBy2 = VAy2 + 2(ay) * dABy

    VBy = VAy + ay * change in time

    Average velocity = dABy / change in time

    3. The attempt at a solution
    VBy2 = VAy2 + 2(ay) * dABy
    (0)2 = (30)2 + 2(9.8) * dABy
    0 = 900 + 19.6 * daby
    -900 = 19.6 *dABy
    -45.9 = dABy
    ^but dABy can't be negative. what am i doing wrong?

    also-- do i use "5 sec" for my time interval on the other equations? i cannot figure out what it means by "first 5 seconds." HELP
     
  2. jcsd
  3. Dec 14, 2009 #2

    PhanthomJay

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    the problem asks for average speed, which is not the same as average velocity
    once you make your minus sign correction, 45.9 m is the distance traveled up. How long did it take to reach that distance? Where will the ball be after 5 seconds? What total distance will the ball have travelled in that 5 seconds (45.9 meters up and an additional distance down)? Then average speed is ???
     
  4. Dec 14, 2009 #3
    well if it travels a total of 2*45.9m ... then the dABy = 91.8

    so:
    average velocity = dABy / t
    = 91.8/5
    =18.36 m/sec
    ???
     
  5. Dec 14, 2009 #4

    PhanthomJay

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    No, first off, you are looking for average speed, after 5 seconds. You can use y = V_yo(t) - 1/2gt^2 (where t=5) to find out the position of the ball after 5 seconds ( it'll be on its descent at that point). Then calculate the average speed by first calculating the total distance it has traveled during that time.
     
  6. Dec 14, 2009 #5
    Thanks! so,
    y =V_yo(t) - 0.5gt^2
    y = 30(5) - 0.5 (9.8) (5)^2
    y = 150 - 4.9 * 25
    y = 150 - 122.5
    y = 27.5 m

    then...
    Vavg = dABy / t
    Vavg = 27.5 m /5 sec
    Vavg = 5.5 m/sec
     
  7. Dec 14, 2009 #6

    PhanthomJay

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    You are getting close, but again, the problem is asking for average speed, not average velocity. What you have calculated is the average velocity, which is displacement divided by time, a vector quantity. Average speed is distance traveled divided by time, a scalar quantity. It travels 45.9 m up and then (45.9 - 27.5) m down, for a total distance of ____? and an average speed of _______?
     
  8. Dec 14, 2009 #7
    Total distance: 45.9 + 23.4 = 63.9 m

    avg speed = dist/time
    avg speed = 63.9 m / 5 sec
    avg speed = 13.9 m/sec^2
     
  9. Dec 14, 2009 #8

    ideasrule

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    45.9-27.5 isn't 23.4. Also, why's your average speed in m/s^2 instead of m/s?
     
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