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Two motion issue

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data
    A boat is 50.0m from the base of a cliff, fleeing at 5.0m/s. A gun, mounted on the edge of the cliff fires a shell at 40.0m/s and hits the boat when it has fled another 50.0m.

    a) At what angel above the horizontal must the gun be aimed so that the shell will hit the target?
    b) How high is the cliff?
    c) With what velocity does the shell hit the boat?

    v = 5m/s
    d = 50m
    t = 10s

    vi = 40m/s
    a = 9.8m/s2
    t = 10s

    2. Relevant equations
    All the ones dealing with motion

    3. The attempt at a solution
    vix = 100/10
    vix = 10m/s

    SinΘ = o/h
    SinΘ = 10/40
    SinΘ = 0.25
    Θ = 14o

    Therefore the angle about the horizontal must be 14o to hit the boat.

    vyf = vyi+aΔt
    vyf = 38.7298+(10x9.8)
    vyf = 38.7298+98
    vyf = 136.73m/s

    Δy = ((136.73+38.7298)/2)x10
    Δy = 87.7x10
    Δy = 877m
    Δy = 880m

    Therefore the cliff is 880m high.


    vf2 = 38.722+(2x9.8x880)
    vf2 = 1500+17195
    vf2 = 18695
    vf = 136.73m/s
    vf = 140m/s

    x2 = 136.732+100
    x2 = 19700
    x = 140.357m/s
    x = 140m/s

    TanΘ = 140/10
    TanΘ = 13.673
    Θ = 85.818o
    Θ = 86o

    Therefore the shell hit the boat at 140m/s [86o below the horizontal]

    Are these answers correct? I find the questions when two objects move at the same time very hard to do.
  2. jcsd
  3. Feb 20, 2010 #2
    What is theta here? As far as I see vx=v*cos(theta)? [part a]
  4. Feb 21, 2010 #3
    I got the angle from using sine. You do not need to use cosine because you have two sides and are looking for an angle. The angle I got was around 14o.
  5. Feb 22, 2010 #4
    I don't get it. A diagram or something?
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