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Homework Help: Sniper Projectile Motion

  1. Jan 28, 2013 #1
    A sniper fires a bullet at 120 m/s at 30° above the horizontal from the roof top of a 35 m high parking garage. If the bullet strikes the level ground beside the parking garage:

    How long was the bullet in the air?

    How far from the base of the parking garage did the bullet land?

    At what angle did the bullet land?

    g= -9.8 m/s^2
    Vi= 120 m/s @ 30 degree incline
    h=35 m
    Vx= 120Cos30 = 140 m/s
    Vy= 120Sin30 = 60 m/s
  2. jcsd
  3. Jan 28, 2013 #2
    Use the following equation to find time:
    Δy = v_oyt + 1/2at^2,
    where Δy = - 35, v_0y = 60, a = -9.8m/s^2
  4. Jan 28, 2013 #3
    First off, doesn't a Muzzle Velocity seem a bit low at 120 m/s? -_-

    Alright, so it looks like I end up with a Quadratic and solve for both values.

    0 = -4.9t^2 + 60t + 35

    t = -.55s and 12.8s

    But there can't be negative time, amiright ?

    And I am unsure about the Total change in the horizontal distance.

    Would ~ Dx = Vx(t) ? [x represents the change in the horizontal or x-axis]

    = (104 m/s)12.8 s
    = 1,331.2 m
  5. Jan 28, 2013 #4
    Ok good, I'm assuming you did the quadratic right.
    If t is 12.8 s.
    Simply use the same equation again:
    Δx = v_oxt + 1/2at^2,
    Here a = 0, so the equation becomes:
    delta x = v_oxt
    You have both of those.
    It seems like you've done it right.
    Good job.
  6. Jan 28, 2013 #5


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    Science Advisor

    It's uses a spring to fire the bullet!

    The point was to separate horizontal (x) and vertical (y) components.

    With straight line constant acceleration a, velocity is [itex]at+ v_0[/itex] and distance traveled is [itex](1/2)at^2+ v_0t+ s_0[itex]

    In this problem, vertically, the acceleration is -g= -9.82 m/s^2, approximately, so that [itex]y= -4.91t^2+ 60t+ 35[/itex] where I have chosen the "0" height to be on the ground, at the bottom of the building. The bullet hits the ground, then, when [itex]-4.91t^2+ 60t+ 35= 0[/itex]. Yes, that will have two roots, one negative and one positive. Yes, the positive root is the one you want.

    Horizontally, the acceleration is 0 so the equation is [itex]x= 140t[/itex] where I have chosen x= 0 as the starting point, horizontally. To find how far the bullet goes horizontally, set t equal to the value you got before for the time the bullet is in the air.
  7. Jan 28, 2013 #6
    Alright, so assuming everything so far is correct, my only issue has been the angle at which the bullet hits the ground.

    If Tan θ = O / A then could I use the components for Sin and Cos ?

    My text says that I should change the Velocity of the Vertical Component by using:

    Vr = Vy + aΔt
    = 60 m/s + (-9.8 m/s)12.8s
    = -65.5 m/s

    Since this becomes the Hypotenuse of the triangle at the moment of impact it becomes positive.

    ... so Tan^-1(104/65.5) = 57.79° from the Vertical.

    Can someone double check if this is an acceptable answer?

    Thanks in advance.
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