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Projectile Motion

  1. May 19, 2009 #1
    1. The problem statement, all variables and given/known data
    There are four different components to this problem. Each component is short -- so don't worry.

    a. Suppose you throw a 0.4 kg ball straight upwards from a height of 1.5 m, at a speed of 4.0 m/sec. How much work have you done on the ball by accelerating it from rest to 4.0 m/s. Is this work positive or negative?

    b. How high do you expect the ball to get above the ground, if there is no air resistance?

    c. At what speed do you expect the ball in #5 to be going when it comes back to the height at which you threw it (1.5 m)?

    d. When the ball comes back down to 1.5 m, you find that it is moving at 3.6 m/sec. You correctly surmise that you can't neglect air resistance here. How much work has "air resistance" done on the ball?


    2. Relevant equations

    x = x0 + voxt + (1/2)(axt2
    v2x = v20x + 2ax (x - x0)
    vx = vox + axt
    WTOT = K2 - K1


    3. The attempt at a solution

    a. WTOT = K2 - K1
    (1/2)(m)(v2) = K
    (1/2)(0.4)(4.0) = 3.2 J = K2.
    K1 = 0 (initial velocity is zero)

    WTOT = 3.2 J

    b. vx = vox + axt
    0 = (4.0) + (-9.8)(t)
    -4.0 = (-9.8)(t)
    -4.0 / -9.8 = t = .408

    My thinking here is that when the ball reaches its maximum height, the velocity is zero.

    x = x0 + voxt + (1/2)(axt2
    x = 1.5 + (4.0)(.408) + (1/2)(-9.8)(.408)2
    x = 2.3163

    c. I'm pretty sure this is just plain wrong. Not sure why, but I have that feeling.

    x = x0 + voxt + (1/2)(axt2
    1.5 = 2.3163 + 0 + (1/2)(9.8)(t2)
    t = .408 (time to go up that distance is the same as the time to go down??)

    vx = vox + axt
    vx = 0 + (9.8)(.408)2 = 1.63 m/s

    d. I know I've screwed up already before. If the ball is moving at 3.6 m/s when it is coming back down WITH air resistance, thats already a bad sign.
     
  2. jcsd
  3. May 19, 2009 #2
    You shouldn't have squared the time there. Intuition would also tell you it will come back down at 4ms at that point-1
     
  4. May 19, 2009 #3

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    (a) and (b) look good. And as Jiacao said, don't square the 0.408 s for part (c), and you'll be in good shape there.

    FYI, for both (b) and (c) you could have used
    v2 = v02 + 2a(x - x0)​

    Then you don't need t.
     
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