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Projectile motion of playing catch

  1. Aug 25, 2013 #1
    1. The problem statement, all variables and given/known data

    You are playing catch with a friend in the hallway of your dormitory. The distance from the floor to the ceiling is D, and you throw the ball with an initial speed v0=√(6gD). What is the maximum horizontal distance (in terms of D) that the ball can travel without bouncing? (Assume the ball is launched from the floor).

    2. Relevant equations

    Range = [v02*sin(2θ)] / g

    y = y0 + xtan(θ)-1/2gx2/(v02*cos2(θ))

    3. The attempt at a solution

    I tried to use the equations and solve for x, without much luck. All my calculations ended up with alot of unknows...
     
    Last edited: Aug 26, 2013
  2. jcsd
  3. Aug 25, 2013 #2

    SteamKing

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    You are given v0, y0 and you know that y <= D. g is a constant. What other unknowns did you come up with? It's very hard to check your work if you don't supply it for review.

    P.S.: What is x supposed to be? Shouldn't you have a variable for time somewhere?
     
  4. Aug 25, 2013 #3

    haruspex

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    Looks like Daniel has substituted for t using t = x / v0 cos(θ). But it has been done incorrectly in the v0 sin(θ) t term, resulting in v0 tan(θ) instead of x tan(θ).

    Daniel, think about the fact that the ball must just avoid hitting the ceiling. What does that tell you about v0 sin(θ)?
     
  5. Aug 26, 2013 #4
    #Haruspex

    Thanks for the correction. It is corrected now.
     
  6. Aug 26, 2013 #5
    Ok so I tried this:

    The max. height is:

    D = [(6gD)*sin(θ)] / (2g)

    which simplifies to:

    D = 3Dsin(θ)

    Hence: sin(θ) = 1/3.

    I tried using this for the horizontal distance:

    R = [(6gD) * (2 cos(θ) sin(θ))] / g

    I found cos(θ) by: √(12-(1/3)2) = (√8)/3

    Next,

    R = [ 6gD * 2 * (√8)/3 * 1/3*] / g = [(8√2)/3]D

    But this answer is incorrect according to the answers which says R = (4√2)D.

    Can you see what I've done wrong?
     
  7. Aug 26, 2013 #6

    CAF123

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    Your method is good, but check this equation.
     
  8. Aug 26, 2013 #7
    #CAF123

    OF COURSE! It is the square of the y-component of the initial velocity! So it's sin squared theta. Thanks for pointing it out and letting me think my self :-).
     
    Last edited: Aug 26, 2013
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