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How far bullet travels in 2,5 seconds

  1. Feb 1, 2015 #1
    I am solving some problems of my physics book for fun and for some reason I am getting a wrong answer from this one.

    1. The problem statement, all variables and given/known data

    A bullet is shot at an angle of 35o with a speed of 90 m/s. I assume that air resistance is ignored.

    Calculate bullet's distance from launch site 2,5 seconds after the shot.

    2. Relevant equations

    x = vox * t
    vox = vo * cosθ

    3. The attempt at a solution

    The horizontal speed should stay as it is.

    vo = 90 m/s
    t = 2,5 s
    θ = 35o

    x = vox * t = vo * cosθ * t = 90 m/s * cos 35o * 2,5 s = 184,309... m ≈ 184 m

    The book says that the answer should be 210 m. I cannot think of any other way to solve this.
     
  2. jcsd
  3. Feb 1, 2015 #2
    The question asked for distance from the launch site, not just the horizontal distance. After 2.5 seconds, the bullet is still in the air, so the vertical distance is nonzero.
     
  4. Feb 1, 2015 #3

    gneill

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    Staff: Mentor

    Has the bullet landed by time t = 2.5 s? If not, what are its coordinates with respect to the launch point?
     
  5. Feb 1, 2015 #4
    Okay, thanks this helped!

    I calculated the vertical distance, y:

    y = voy * t - 0,5 * g * t2
    = vo * sinθ * t - 0,5 * g * t2
    = 90 m/s * sin 35o * 2,5 s - 0,5 * 9,81 m/s2 * (2,5 s)2
    = 98,398...m

    Then I figured that the real distance, d, is the hypotenuse of x and y.

    d = √(x2 + y2) = √((184,309... m)2 + (98,398... m)2)
    = 208,930... m
    ≈ 210 m

    This should be correct way to do it, right?
     
  6. Feb 1, 2015 #5

    PeroK

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    Science Advisor
    Homework Helper
    Gold Member

    Right!
     
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