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Please Help Me Find Horizontal Distance?

  1. Sep 6, 2011 #1
    A ball is thrown with an initial speed of 4.6m/s at an angle of 15(degrees) below the horizontal. It is released 0.80m above the floor.

    What horizontal distance does the ball cover before bouncing?



    I keep getting 2.4m as my answer, but it's wrong!
    Here's what I did (Please correct any mistakes):


    4.6sin(15) = initial vertical velocity = 1.19m/s

    1.19^2 = 2gh

    1.42 = 2gh

    1.42/(2*9.81)=h= .072m

    upward time = 1.19 m/s - gt = 0
    1.19/9.81 = t = .121sec

    total height = .072 + 0.8 = .872m

    0.872 = 1/2gt^2
    0.872/4.9 = t^2 = 0.178, sqrt = t = 0.422sec

    total time = 0.121 + 0.422= 0.543sec

    horizontal velocity = 4.6cos(15) = 4.44m/s

    4.44x 0.543 = 2.41m (answer)
     
  2. jcsd
  3. Sep 7, 2011 #2

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  4. Sep 7, 2011 #3
    Remember that gravity acts on the object as soon as it is released, so the time it is in the air is dependent on that, regardless of how fast you're throwing it in the horizontal direction.
     
    Last edited: Sep 7, 2011
  5. Sep 7, 2011 #4
    A ball is thrown with an initial speed of 4.6m/s at an angle of 15(degrees) below the horizontal. It is released 0.80m above the floor.

    What horizontal distance does the ball cover before bouncing?

    Is the answer 2.98?

    Here's what I did:
    4.6sin(15) = initial vertical velocity = 1.19m/s
    horizontal velocity = 4.6cos(15) = 4.44m/s


    .8/(1.19m/s)=.672s
    0+(4.44m/s)(.672s)=2.98
     
  6. Sep 7, 2011 #5

    tiny-tim

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    hi kman2027! :smile:
    the first equation assumes that the vertical speed is constant (s = ut), it isn't :redface:

    you need to use one of the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations to find t

    (and then your method in the second equation is correct, since the horizontal speed is constant :wink:)
     
    Last edited by a moderator: Apr 26, 2017
  7. Sep 7, 2011 #6
    I got 1.7m this time.

    I picked the constant acceleration equation:

    y=yo+voyt-1/2gt^2
    plugged in 0=.8+1.19t-1/2(-9.8)t^2
    t=.385

    (.385)(4.44)=1.7m

    Is 1.7m right?
     
  8. Sep 8, 2011 #7

    tiny-tim

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    hi kman2027! :smile:

    (just got up :zzz: …)
    not quite …

    the initial vertical speed is negative :wink:

    (and you shouldn't have both those minuses in the t2 term)
     
  9. Sep 8, 2011 #8
    My bad, I had a typo in the equation.

    .8-1.19t+(1/2)(-9.8)t^2=0
    t= .3s

    Then:
    (4.44)(.3)= 1.3m

    Is 1.3m it?

    Thanks for your help by the way!
     
  10. Sep 8, 2011 #9
    What's the equation for calculating distance based on velocity and time?
     
  11. Sep 8, 2011 #10

    tiny-tim

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    yes 1.3 looks fine :smile:
     
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