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Bowling Ball Thrown off Building-Check my work/solutions pls.

  1. Feb 4, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball rolls down a roof that makes an angle of 30 degrees to the horizontal. It rolls
    off the edge woth speed of 5.00 m/s. The distance to the ground from that point is 7.00 m.
    a) How long is the ball in the air for?
    b) How far from the base of the house does it land?
    c) What is its speed before landing?

    2. Relevant equations
    [tex]\vartheta[/tex] = 30 degrees
    speed = 5.00 m/s
    distance from ground to point ball being dropped (roof) = 7.00 m

    3. The attempt at a solution
    a) time in the air:
    y = -7.00m
    vx0 =5.00 m/s
    ax = 0 <- drop so acceleration is 0..correct?
    vy0 = 0
    ay = -g <---gravity
    x0 and y0 = 0

    y = - 1/2 gt2
    t = sqrt [(2y)/(-g) = sqrt [(2)(-7.00)/(-9.80) = 1.2 s <--ball in the air for

    b) far from the base of the house:
    x = vx0t = (5.00)(1.2) = 6 m<---- distance from base of the house

    c) i dont know how to find the speed-do i need the angle in this equation?
  2. jcsd
  3. Feb 4, 2009 #2


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    I don't think part a is correct.
    In part (a), you are only concerned with the vertical motion so the distance is 7 m as you have, but the initial velocity is only 5*sin(30). The initial velocity is not zero so you must use the whole formula for distance: y = Vo*t + .5*a*t^2.
  4. Feb 4, 2009 #3
    but then i get -2.88 <---since a was -g (-9.80)

    initial velocity was: 2.5
    and then y came out to be -4.06.

    is that correct? is my base distance correct??
  5. Feb 4, 2009 #4


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    I don't get 2.88. Isn't y = 7?
    What formula are you using?
  6. Feb 4, 2009 #5
    this is what i used:
    y = Vo*t + .5*a*t^2.

    ^unless you are talking about initial y!
  7. Feb 4, 2009 #6


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    y = Vo*t + .5*a*t^2.
    7 = ???
  8. Feb 4, 2009 #7
    7 = (5 sin 30)(1.2) + .5(-9.80)(1.2)^2
    v0 = 5 sin 30
    t = 1.2 s
    a = -g = -9.80
    7 = 3 + (-7.056)
    7 = -4.056 <---that doesnt make sense, am i suppose to solve for a variable?
  9. Feb 4, 2009 #8


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    "7 = (5 sin 30)(1.2) + .5(-9.80)(1.2)^2"
    Okay - now I can see what you are doing!
    First, you don't know what the time is, so you shouldn't be putting in 1.2 for t.
    In fact, we are trying to find the time, so it must appear as a t.
    Second, down is positive in this problem, so drop the minus sign on the 9.8.

    Then you are ready to solve the formula for time. Looks like it is a quadratic!
  10. Feb 4, 2009 #9

    i thought i solved for time correctly-i got 1.2s for t.
    so it is:
    7 = (5 sin 30)(t) + .5(9.80)(t)^2
    7 = 2.5t + 4.9t^2
    0 = 4.9t^2 + 2.5t -7
    and time should be in the positive correct?
  11. Feb 4, 2009 #10
    now how do i find the speed??
  12. Feb 4, 2009 #11


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    Yes, looks correct. What time did you find?
    Speed should be done in two parts - vertical and horizontal.
    Horizontal is easy - no acceleration.
    Vertically not bad when you know the initial speed and the time of fall.
  13. Feb 4, 2009 #12
    im using the quad formula which i assume is the way to go.
    but my answers are coming in the neg. im trying to rework it.

    what do you mean by vertical and horizontal? how do i find that? (sorry its my first time takin physics) you mentioned it in my other prblm as well so i want to make sure im doin it rite.

    by the way-thanks for helpin me with my prblms! it means a lot! :D
  14. Feb 4, 2009 #13
    ok so here is my solution for t:
    used quad formula
    (-2.5 + 12) / 9.8 = 0.97 s
  15. Feb 4, 2009 #14


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    The quadratic formula gives one negative and one positive answer. Ignore the negative - something that happened before the problem started!

    You have 5 m/s at 30 degrees below horizontal. Make a triangle around that with a horizontal side and a vertical side. The vertical side has length 5 sin(30). The horizontal side has length 5*cos(30). That is your horizontal velocity.

    Use V = Vo + at to get the vertical velocity at the time it reaches the ground.

    Finally, you must combine the horizontal and final vertical velocities using a triangle again. This time you must calculate the hypotenuse to get the combined velocity.

    Most welcome - thanks for using PF.
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