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Help getting projectile motion equations straight in my head

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data
    In the figure, you throw a ball toward a wall at speed 26.0 m/s and at angle θ0 = 37.0˚ above the horizontal. The wall is distance d = 17.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall?

    [itex]
    \theta _0 = 37^o
    [/itex]

    [itex]
    v _0 = 26 \frac{m}{s}
    [/itex]

    [itex]
    (x - x_0) = \Delta x = 17 m
    [/itex]

    2. Relevant equations

    [itex]
    \Delta x = (v_0 cos \theta_0) t
    [/itex]

    [itex]
    t = \frac{\Delta x}{(v_0 cos \theta_0)} \approx 0.8187
    [/itex]

    [itex]
    \vec{V}_x = v_0 \cos (\theta_0)
    [/itex]

    [itex]
    \vec{V}_y = v_0 \sin (\theta_0) - gt
    [/itex]

    3. The attempt at a solution

    So I ended up finishing this problem, but I'm having trouble with these equations and the way they're written.

    It's my understanding that acceleration (a) has direction. When a = gravity (g) it's value is -9.8 m/s2. However when Halliday writes in the textbook, he writes the equations this way:

    [itex]
    (y-y_0) = (v_0 \sin \theta_0)t - \frac{1}{2}g t^2
    [/itex]

    So is this an error in the text? The -g is redundant since gravity is accelerating objects toward the x-axis?
     
    Last edited: Sep 18, 2011
  2. jcsd
  3. Sep 18, 2011 #2
    What you have looks good, the way most coordinate systems are set up you want a minus sign for the term gt^2/2, as you have it.
     
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