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Projectile Motion of a snowball

  1. Nov 7, 2007 #1
    Q: A child throws a snowball with a horizontal velocity of 18 m/s directly toward a tree,
    from a distance of 9.0 m and a height above the ground of 1.5 m.

    a) After what interval does the snowball hit the tree?
    b) At what height above the ground will the snowball hit the tree?
    c) Determine the snowball's velocity as it strikes the tree.

    I have done a) and c) but I cannot do b).
    I know that when the snowball hits the tree the velocity will be zero,
    so Vf = 0, and it happens at 1.8 so I know that t =1.8 but I have no clue
    what to do next. Which equation should I use?

    Thanks in advance
     
  2. jcsd
  3. Nov 7, 2007 #2

    mjsd

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    may be it is a good idea that you show us the work you have done and we can then point out to you when you have gone wrong (if there are actually any mistakes)
     
  4. Nov 7, 2007 #3
    I just typed the question right out of the work book so 1.5m above the ground
    is the only height I got in this question :(

    a) since i know that d[tex]_{y}[/tex] = 1.5m and a = -9.8m/s[tex]^{2}[/tex]
    i substituted into the equation, d[tex]_{y}[/tex] = v[tex]_{iy}[/tex][tex]\Delta[/tex]t + 1/2a[tex]\Delta[/tex]t[tex]^{2}[/tex],
    v[tex]_{iy}[/tex], which is the initial vertical velocity, it is zero, so i end up getting 1.8 for t. therefore, the snowball will hit the tree at 1.8 sec
     
  5. Nov 7, 2007 #4

    mjsd

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    yeah, I realised that after I posted

    that's only tell you how long it would take to hit the ground (if the tree is not there)!
     
  6. Nov 7, 2007 #5
    for c), i know that v[tex]_{fx}[/tex] = v[tex]_{ix}[/tex] = 18m/s, therefore,
    subsitute into v[tex]_{fy}[/tex] = v[tex]_{iy}[/tex] + a x t, since v[tex]_{iy}[/tex] = 0,
    and a = -9.8m/s[tex]^{2}[/tex], i get;
    v[tex]_{fy}[/tex] = -17.64

    v[tex]_{f}[/tex] = (18)[tex]^{2}[/tex] + (-17.64)[tex]^{2}[/tex]
    v[tex]_{f}[/tex] = 25m/s

    tan[tex]\theta[/tex] = 17.64 / 18
    [tex]\theta[/tex] = 44 degrees.

    Therefore, the velocity is 25 m/s @ 44 degrees above surface
     
  7. Nov 7, 2007 #6
    uh oh then i'm in a big trouble lol
    so I have to find how tall the tree is?
     
  8. Nov 7, 2007 #7

    mjsd

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    do you realise so far you haven't used the info about 9m from tree? :smile:
     
  9. Nov 7, 2007 #8
    oh oh, so i've used a different equation and i get a different t value,

    d[tex]_{x}[/tex] = v[tex]_{ix}[/tex] t + 1/2 a t[tex]^{2}[/tex]
    9.0m = (18m/s)(t) + 1/2(-9.8)(t[tex]^{2}[/tex])
    9.0m = t(18m/s - 4.9m/s[tex]^{2}[/tex] t)
    t = 0, OR
    4.9m/s t = 18m/s
    t = 3.6 seconds

    would this be a new answer, or did i just do something stupid? lol
     
  10. Nov 7, 2007 #9

    mjsd

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    by the way, there is NO acceleration/deceleration in the x direction...i don't think...
     
  11. Nov 7, 2007 #10
    oh you're right.. my bad,

    then i'm really stuck..
    if my first answer was only telling "how long it will take for the snowball to hit ground",
    how can i find the tree's height and relate this equation to it?
     
  12. Nov 7, 2007 #11

    mjsd

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    you don't need tree height. all you need is how long it will take to hit tree (x-dir) and then that time will tell you how far it has dropped during that time (y-dir), and eventually you can get velocity in x and y then do a vector sum.
     
  13. Nov 7, 2007 #12
    i'm sorry but i have no clue,
    i thought my answer to a) was right but since it's wrong
    i don't know where to start anymore..
     
  14. Nov 8, 2007 #13

    mjsd

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    hint: velocity in x direction will not change since there is no acceleration in that direction and then u r given the distance (9m) plus initial velocity in x dir,.. so how do you find the time it takes to reach the tree ?
     
  15. Nov 8, 2007 #14
    d[tex]_{x}[/tex] = v[tex]_{x}[/tex] x t
    t = 9.0 m / 18 m/s
    t = 0.5 s

    is this wrong?
     
  16. Nov 8, 2007 #15

    mjsd

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    sounds good, now check that and see how far it have dropped in that time and hopefully it is still above ground! otherwise there is a contradiction :smile:
     
  17. Nov 8, 2007 #16
    Thank You,
    So I used this equation to find the height at t = 0.5, [tex]\Delta[/tex]d[tex]^{}_{y}[/tex] = v[tex]^{}_{iy}[/tex][tex]\Delta[/tex]t + 1/2a[tex]\Delta[/tex]t[tex]^{2}[/tex]
    since v[tex]^{}_{iy}[/tex] = 0,
    [tex]\Delta[/tex]d[tex]^{}_{y}[/tex] = 1/2(-9.8m/s[tex]^{2}[/tex])(0.5)[tex]^{2}[/tex]
    [tex]\Delta[/tex]d[tex]^{}_{y}[/tex] = -1.2 m

    I get negative answer :(
    Did I use a wrong formula?
     
  18. Nov 9, 2007 #17

    mjsd

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    formula is ok, coordinate system used has been confused (by youself)
    in setting a = -9.8m/s^2, you have basically chosen the positive y direction to be upwards.
    so your displacement of -1.2m is of course meaning 1.2m downwards from starting point.
    now, you should be able to finish the entire problem without any more of my help.
    cheers
     
  19. Nov 9, 2007 #18
    oooh so since the kid is located 1.5m up, it would be 1.5 -1.2 = 30cm
    and thats the height the tree was hit with the snowball.

    i get it now!
    thanks for the looong two days of help :)
     
  20. Jan 6, 2011 #19
    Is this correct?
     
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