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Homework Help: 2D kinematics - find angle

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A sniper barrel is exactly one meter above the ground and perfectly horizontal. 1000m away is a target one meter above the ground. if the bullet leaves the gun with a muzzle velocity of 1500m/s will it reach the target? (no, next part is my problem) at what angle should the sniper shoot to hit the target?

    [tex]\Delta[/tex]x = 1000m
    [tex]\Delta[/tex]y = 0m
    ax = 0
    ay = -9.8m/s2 (gravity)
    V0x = ?
    V0y = ?
    time (t) = ?
    angle [tex]\theta[/tex] = ?

    V0 = 1500m/s

    2. Relevant equations
    [tex]\Delta[/tex]X = v*t
    [tex]\Delta[/tex]V = V - V0 = a*t
    [tex]\Delta[/tex]X = (1/2)*a*t2 + V0*t
    V2 = 2*a*[tex]\Delta[/tex]X - V02

    V0x = V0*cos[tex]\Theta[/tex]
    V0y = V0*sin[tex]\Theta[/tex]

    *the first four equations can be used for either the X or Y dimension; [tex]\Delta[/tex]X will just change to [tex]\Delta[/tex]Y

    3. The attempt at a solution

    I'm not looking for the answer, just a point in the right direction please.

    For part one of the question, I found that it would take the bullet 2/3 of a second to reach the target in the X dimension. But in that same time, it would fall 2.2m, hitting the ground 322.3m from the target.
    Last edited: Mar 22, 2009
  2. jcsd
  3. Mar 22, 2009 #2


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  4. Mar 22, 2009 #3
    Thanks for the reply. I checked out that link and then fixed my original post a bit, putting in the equations as I learned them.
  5. Mar 22, 2009 #4


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    Now you need to figure the time it will be in the air as a function of the vertical velocity. Up and down.

    Then figure the same time as a function of the horizontal velocity and the known distance. These times of course must be equal, so that begins to put you in position to figure out what angle satisfies both conditions.
  6. Mar 23, 2009 #5
    I've done that and get:

    (-2V0y)/(-9.8) = Time-y
    1000/V0x = Time-x

    then to

    ((-9.8)1000)/V0x = -2V0y

    ((-9.8)1000)/(1500cos[tex]\Theta[/tex]) = -2(1500)sin[tex]\Theta[/tex]

    ((-9.8)1000) = -2(1500)sin[tex]\Theta[/tex](1500cos[tex]\Theta[/tex])

    ((-9.8)1000) = (-2)(1500)(1500)sin[tex]\Theta[/tex]cos[tex]\Theta[/tex]

    ((-9.8)(1000))/((-2)(1500)(1500)) = sin[tex]\Theta[/tex]cos[tex]\Theta[/tex]

    0.00217777778 = sin[tex]\Theta[/tex]cos[tex]\Theta[/tex]

    Now what?
  7. Mar 23, 2009 #6


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    You might want to recognize the identity

    2SinθCosθ = Sin2θ

    It's a terrifically useful trig identity that you may run into again with projectile motion equations.
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